The area of the rectangle is the product of the length and the width of the rectangle. The area is also the sum of the subareas:

a.
$$
(x+4)(x+y+1)=xy+4y+x^2+6x+8
$$

b.
$$
(6x-1)(3x+2)=18x^2+12x-3x-2=18x^2+9x-2
$$

a. $(x+4)(x+y+1)=xy+4y+x^2+6x+8$

b. $(6x-1)(3x+2)=18x^2+12x-3x-2=18x^2+9x-2$

No, it is not possible to factor every expression.

The area of the rectangle is the product of the length and the width of the rectangle. The area is also the sum of the subareas:

a.
$$
2x^2+7x+6=(2x+3)(x+2)
$$

b.
$$
6x^2+7x+2=(3x+2)(2x+1)
$$

c. Not possible, since the first term is $x^2$ both factor should contain $x$ and since the last term is 1 both factors should contain 1, but as you can see in the area model we then obtain the polynomial $x^2+2x+1$.

d.
$$
2xy+y^2+6x+3y=(2x+y)(y+3)
$$

a. $(2x+3)(x+2)$

b. $(3x+2)(2x+1)$

c. Not factorable

d. $(2x+y)(y+3)$

The area of the rectangle is the product of the length and the width of the rectangle. The area is also the sum of the subareas:

a.
$$
(1+3x)(2x+5)=6x^2+17x+5
$$

b.
$$
(-2+5x)(y+3)=-2y-6+5xy+15x
$$

c.
$$
(-3+4x)(3x+4)=12x^2+7x-12
$$

Strategy: find the greatest common factors in every row and column and write these at the beginning of the row or the bottom of the column.

a. $(1+3x)(2x+5)=6x^2+17x+5$

b. $(-2+5x)(y+3)=-2y-6+5xy+15x$

c. $(-3+4x)(3x+4)=12x^2+7x-12$

a. One diagonal contains the factor $x$ and the other diagonal doesn’t.

b. No, because we note that the the 2-by-2 area model in (b) does not fit this property due to the other variable $y$ being present.

a. One diagonal contains the factor $x$ and the other diagonal doesn’t.

b. No

The area of the rectangle is the product of the length and the width of the rectangle. The area is also the sum of the subareas:

a.
$$
(2x-3)(4x+1)=8x^2-12x+2x-3=8x^2-10x-3
$$

b.
$$
(4x-8)(4x-8)=16x^2-32x-32x+64=16x^2-64x+64
$$

a. $8x^2-10x-3$

b. $16x^2-64x+64$

The area of the rectangle is the product of the length and the width of the rectangle. The area is also the sum of the subareas:

$$
(-1+2x)(x+3y-5)=-11x-3y+5+2x^2+6xy
$$

$$
(-1+2x)(x+3y-5)=-11x-3y+5+2x^2+6xy
$$

a. The greatest common factor is 4:

$$
4(x+2)
$$

b. The greatest common factor is 5:

$$
5(2x+5y+1)
$$

c. The greatest common factor is $2x$:

$$
2x(x-4)
$$

d. The greatest common factor is $3x$:

$$
3x(3xy+4+y)
$$

a. $4(x+2)$

b. $5(2x+5y+1)$

c. $2x(x-4)$

d. $3x(3xy+4+y)$

The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
tan{17text{textdegree}}=dfrac{9.0}{x}
$$

Multiply both sides of the equation by $x$:

$$
xtan{17text{textdegree}}=9.0
$$

Divide both sides of the equation by $tan{17text{textdegree}}$:

$$
x=dfrac{9.0}{tan{17text{textdegree}}}approx 29.4cm
$$

The hypotenuse can then be calculated by the Pythagorean theorem:

$$
c=sqrt{9.0^2+29.4^2}=sqrt{945.36}approx 30.7cm
$$

The perimeter is then the sum of all sides:

$$
9.0cm+30.7cm+29.4cm=69.1cm
$$

a. Function, because every $x$-value has exactly one corresponding $y$-value.

b. No function, because if you would draw a vertical line, then the graph will intersect it twice.

c. Function, because every $x$-value has exactly one corresponding $y$-value.

d. Function, because every $x$-value has exactly one corresponding $y$-value.

a. The probability of an event not happening is 100% decreased by the probability of the event happening:

$$
P(text{ student is female })=100%-P(male)=100%-25%=75%
$$

b. The probability of both events is the product of the probability of each event:

$$
P(maletext{ and } not walk)=P(male)cdot P(not walk)=25% cdot (100%-40%)=0.25cdot 0.6=0.15=15%
$$

c. Addition rule: $P(Atext{ or }B)=P(A)+P(B)-P(Atext{ and }B)$:

$$
P(walktext{ or } not walk)=P(walk)+P(not walk)-P(walktext{ and } not walk)=40%+60%-0%=100%
$$

d. (b) uses AND and is thus an intersection, (c) uses OR and is thus a union.

The area of the rectangle is the product of the length and the width of the rectangle. The area is also the sum of the subareas:

a.
$$
(-7+2x)(5x-2)=10x^2-39x+14
$$

b. The bottom left corner contains the product of two terms and the upper right corner contains the product of two terms.

a. $(-7+2x)(5x-2)=10x^2-39x+14$

b. Yes

**Concept**
A quadratic expression in the form: $ax^2 + bx + c$, can be represented as a product of two binomial expressions as follows:
$$begin{gather}
ax^2 + bx + c &= (dx+e)(fx+g)
end{gather}$$

**Part a**
We are to use algebra tiles to obtain the factors of the expression $2x^2 + 5x +3$. We first obtain the factors of the expression using Eq (1) as follows:
$$begin{aligned}
ax^2 + bx + c &= (dx+e)(fx+g) \
2x^2 +5x+3 &= (2x+3)(x+1) \
2x^2 +5x+3 &= 2x^2 + 3x + 2x + 3 \
2x^2 +5x+3 &= 2x^2 +5x+3
end{aligned}$$

After obtaining and verifying the factors, we then place the values inside each algebra tile as follows:

We the determine the area as a product as follows:
$$begin{aligned}
A_1 &= text{length} times text{width} \
&= (2x + 3) times (x+1) \
&= 2x^2+ 3x + 2x + 3 \
&= boxed{2x^2+5x +3}
end{aligned}$$

**Part b**
We are to use algebra tiles to obtain the factors of the expression $3x^2 + 10x +8$. We first obtain the factors of the expression using Eq (1) as follows:
$$begin{aligned}
ax^2 + bx + c &= (dx+e)(fx+g) \
3x^2 + 10x +8 &= (3x+4)(x+2) \
3x^2 + 10x +8 &= 3x^2 + 4x + 6x + 8 \
3x^2 + 10x +8&= 3x^2 +10x+8
end{aligned}$$

After obtaining and verifying the factors, we then place the values inside each algebra tile as follows:

We then determine the area as a product as follows:
$$begin{aligned}
A_2 &= text{length} times text{width} \
&= (3x+4) times (x+2) \
&= 3x^2+ 4x + 6x + 8 \
&= boxed{3x^2+10x +8}
end{aligned}$$

**Part c**
We are to use algebra tiles to obtain the factors of the expression $2x^2 + 7x +6$. From Casey’s pattern for diagonals, we know that the product of the diagonals should be equal. Hence, we obtain the two values that would be equal to the product of the given diagonal as follows:
$$begin{aligned}
D_1 &= D_2 \
D_1 &= 2x^2 times 6 \
D_1 &= 12x^2 \
D_2 &= 3x times 4x \
D_2 &= 12x^2
end{aligned}$$

Hence, the missing values are $3x$ and $4x$

**Part d**
We know that a diamond problem has the form:

From the values in part (c), we set $x=3x$ and $y=4x$.

**Part e**
We then illustrate the area model using the values from part (c) as follows:

We then determine the area as a product as follows:
$$begin{aligned}
A_4 &= text{length} times text{width} \
&= (2x+3) times (x+2) \
&= 2x^2+ 4x + 3x + 6 \
&= boxed{2x^2+7x +6}
end{aligned}$$

**Given**
From the expression $6x^2 +17x +12$, we obtain the area model using the diamond problem to obtain the factors of the expression.

Using Casey’s pattern we find the missing values by getting the product of the diagonals as follows:

$$begin{aligned}
D_1 &= D_2 \
D_1 &= 6x^2 times 12 \
D_1 &= 72x^2 \
D_2 &= 8x times 9x \
D_2 &= 72x^2
end{aligned}$$

Hence, the missing values are $8x$ and $9x$

We set $x=8x$ and $y=9x$. We then obtain the area model using the diamond problem as follows:

From the obtained values, we know that the values $8x$ and $9x$ should be the product of the factors. Hence, the possible factors of the expression are $(3x+4)$ and $(2x+3)$. We verify these factors as follows:
$$begin{aligned}
6x^2 +17x +12 &= (3x+4) (2x+3) \
6x^2 +17x +12 &= (3x cdot 2x) + 8x + 9x + 12 \
6x^2 +17x +12 &= 6x^2 + 17x + 12
end{aligned}$$

Therefore, the factors $(3x+4)(2x+3)$ yields the expression $6x^2 +17x +12$

**Part a**
Given the trinomial $6x^2 + 17x + 12$, we place the first and last term of the expression in the area model as follows:

We obtain the missing values of the area model by getting the product of the diagonals as follows:

$$begin{aligned}
D_1 &= D_2 \
D_1 &= 6x^2 times 12 \
D_1 &= 72x^2 \
D_2 &= 8x times 9x \
D_2 &= 72x^2
end{aligned}$$

Hence, the missing values in the area model are $8x$ and $9x$

**Part b**
A diamond problem has the form:

The product $(xy)$ goes on the top, while the sum $(x+y)$ goes on the bottom.

**Part c**
We set-up the diamond problem by setting $x=8x$ and $y=9y$. The diamond problem is solved as follows:

**Part d**
Using the area model in part (c), we obtain the area of the rectangle as a product as follows:
$$begin{aligned}
A &= text{length} times text{width} \
&= (3x+4) times (2x+3) \
&= (3x cdot 2x) + 8x + 9x + 12 \
&= boxed{6x^2 + 17x + 12}
end{aligned}$$

**Part a**
Given the trinomial $x^2 + 9x +18$, we place the first and last term of the expression in the area model as follows:

We obtain the missing values of the area model by getting the product of the diagonals as follows:

$$begin{aligned}
D_1 &= D_2 \
D_1 &= x^2 times 18 \
D_1 &= 18x^2 \
D_2 &= 6x times 3x \
D_2 &= 18x^2
end{aligned}$$

Hence, the missing values in the area model are $3x$ and $6x$

Using the values $3x$ and $6x$, the area model can be represented as follows:

We obtain the area of the rectangle be getting the product of the length and width as follows:
$$begin{aligned}
A &= text{length} times text{width} \
&= (x+6) times (x+3) \
&= x^2 + 6x + 3x + 18 \
&= x^2 + 9x+18
end{aligned}$$

Therefore, the factors $boxed{(x+6)(x+3)}$ yields the expression $x^2+9x+18$

**Part b**
Given the trinomial $4x^2 + 17x -15$, we place the first and last term of the expression in the area model as follows:

We obtain the missing values of the area model by getting the product of the diagonals as follows:

$$begin{aligned}
D_1 &= D_2 \
D_1 &= 4x^2 times (-15) \
D_1 &= -60x^2 \
D_2 &= 20x times (-3x) \
D_2 &= -60x^2
end{aligned}$$

Hence, the missing values in the area model are $20x$ and $-3x$

Using the values $20x$ and $-3x$, the area model can be represented as follows:

We obtain the area of the rectangle be getting the product of the length and width as follows:
$$begin{aligned}
A &= text{length} times text{width} \
&= (x+5) times (4x-3) \
&= 4x^2 + 20x – 3x -15 \
&= 4x^2 + 17x-15
end{aligned}$$

Therefore, the factors $boxed{(x+5)(4x-3)}$ yields the expression $4x^2 + 17x-15$

**Part c**
Given the trinomial $4x^2 – 8x +3$, we place the first and last term of the expression in the area model as follows:

We obtain the missing values of the area model by getting the product of the diagonals as follows:


$$begin{aligned}
D_1 &= D_2 \
D_1 &= 4x^2 times 3 \
D_1 &= 12x^2 \
D_2 &= -2x times -6x \
D_2 &= 12x^2
end{aligned}$$

Hence, the missing values in the area model are $-2x$ and $-6x$

Using the values $-2x$ and $-6x$, the area model can be represented as follows:


We obtain the area of the rectangle be getting the product of the length and width as follows:
$$begin{aligned}
A &= text{length} times text{width} \
&= (2x-1) times (2x-3) \
&= 4x^2 – 2x – 6x +3 \
&= 4x^2 – 8x+3
end{aligned}$$

Therefore, the factors $boxed{(2x-1)(2x-3)}$ yields the expression $4x^2 – 8x+3$

**Part d**
Given the trinomial $3x^2 + 5x -3$, we place the first and last term of the expression in the area model as follows:


We obtain the missing values of the area model by getting the product of the diagonals as follows:

$$begin{aligned}
D_1 &= D_2 \
D_1 &= 3x^2 times (-3) \
D_1 &= -9x^2 \
D_2 &= -3x times 3x \
D_2 &= -9x^2
end{aligned}$$

Hence, the missing values in the area model are $-3x$ and $3x$

Using the values $-3x$ and $3x$, the area model can be represented as follows:


We obtain the area of the rectangle be getting the product of the length and width as follows:
$$begin{aligned}
A &= text{length} times text{width} \
&= (x+1) times (x-1) \
&= x^2 +x – x -1 \
&= x^2 – 1
end{aligned}$$

The result obtained is not equal to the original expression.

Using the obtained values $3x$ and $-3x$, we observe that these values cannot be used to obtain the factors of the quadratic equation $3x^2 +5x-3$. Moreover, this quadratic equation cannot be factored using the area model nor diamond problem method.

a. $(x+3)(x+6)$
b. $(4x-3)(x+5)$
c. $(2x-1)(2x-3)$
d. Cannot be factored

Quadratic Form.
$$begin{aligned}
ax^2+bx+c
end{aligned}$$

Example.
$$begin{aligned}
8x^2+2x-45
end{aligned}$$

Let $a=8$, $b=2$ and $c=-45$

Based on table shown in **Step 7**, the two numbers whose product $(-360)$ and sum $(2)$ are $-18$, and $20$.

Using the diamond method shown in **Step 6**, the **factored form** of $8x^2+2x-45$ is,
$$begin{aligned}
(2x+5)(4x-9)
end{aligned}$$

**Part a**
Given the trinomial $x^2 + 9x +18$, we place the first and last term of the expression in the area model as follows:

We obtain the missing values of the area model by getting the product of the diagonals as follows:

$$begin{aligned}
D_1 &= D_2 \
D_1 &= x^2 times (-12) \
D_1 &= -12x^2 \
D_2 &= 2x times (-6x) \
D_2 &= -12x^2
end{aligned}$$

Hence, the missing values in the area model are $2x$ and $-6x$

Using the values $2x$ and $-6x$, the area model can be represented as follows:

We obtain the area of the rectangle be getting the product of the length and width as follows:
$$begin{aligned}
A &= text{length} times text{width} \
&= (x-6) times (x+2) \
&= x^2 – 6x + 2x -12 \
&= x^2 – 4x-12
end{aligned}$$

Therefore, the factors $boxed{(x-6)(x+2)}$ yields the expression $x^2-4x-12$

**Part b**
Given the trinomial $2x^2 – 9x -5$, we place the first and last term of the expression in the area model as follows:

We obtain the missing values of the area model by getting the product of the diagonals as follows:


$$begin{aligned}
D_1 &= D_2 \
D_1 &= 2x^2 times (-5) \
D_1 &= -10x^2 \
D_2 &= x times (-10x) \
D_2 &= -10x^2
end{aligned}$$

Hence, the missing values in the area model are $x$ and $-10x$

Using the values $x$ and $-10x$, the area model can be represented as follows:

We obtain the area of the rectangle be getting the product of the length and width as follows:
$$begin{aligned}
A &= text{length} times text{width} \
&= (x-5) times (2x+1) \
&= 2x^2 – 10x + x -5 \
&= 2x^2 – 9x-5
end{aligned}$$

Therefore, the factors $boxed{(x-5)(2x+1)}$ yields the expression $2x^2-9x-5$

**Part c**
Given the trinomial $4x^2 +4x +1$, we place the first and last term of the expression in the area model as follows:


We obtain the missing values of the area model by getting the product of the diagonals as follows:

$$begin{aligned}
D_1 &= D_2 \
D_1 &= 4x^2 times 1 \
D_1 &= 4x^2 \
D_2 &= 2x times 2x \
D_2 &= 4x^2
end{aligned}$$

Hence, the missing values in the area model are $2x$ and $2x$

Using the values $2x$ and $2x$, the area model can be represented as follows:


We obtain the area of the rectangle be getting the product of the length and width as follows:
$$begin{aligned}
A &= text{length} times text{width} \
&= (2x+1) times (2x+1) \
&= 4x^2 +2x +2x+1 \
&= 4x^2 + 4x+1
end{aligned}$$

Therefore, the factors $boxed{(2x+1)(2x+1)}$ yields the expression $4x^2+4x+1$

**Part d**
Given the trinomial $3x^2 +10x -8$, we place the first and last term of the expression in the area model as follows:

We obtain the missing values of the area model by getting the product of the diagonals as follows:


$$begin{aligned}
D_1 &= D_2 \
D_1 &= 3x^2 times (-8) \
D_1 &= -24x^2 \
D_2 &= 12x times (-2x) \
D_2 &= -24x^2
end{aligned}$$

Hence, the missing values in the area model are $12x$ and $-2x$

Using the values $12x$ and $-2x$, the area model can be represented as follows:


We obtain the area of the rectangle be getting the product of the length and width as follows:
$$begin{aligned}
A &= text{length} times text{width} \
&= (3x-2) times (x+4) \
&= 3x^2 +12x -2x-8 \
&= 3x^2 + 10x-8
end{aligned}$$

Therefore, the factors $boxed{(x+4)(3x-2)}$ yields the expression $3x^2+10x-8$

a. $(x-6)(x+2)$
b. $(2x-1)(x+5)$
c. $(2x+1)(2x+1)$
d. $(3x-2)(x+4)$

The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
tan{68text{textdegree}}=dfrac{x}{15}
$$

Multiply both sides of the equation by $15$:

$$
37.1approx 15tan{68text{textdegree}}=x
$$

Add the height of Mary’s eyes with respect to the ground to the found height:

$$
37.1’+32’=69.1′
$$

b. In this case, the graph has a minimum at $(0,0)$ and thus this is the vertex.

Vertex $(0,0)$ is minimum

The expected value is the sum of the products of the possibilities and their probability:

a.
$$
EV=9cdot dfrac{1}{12}+4cdot dfrac{11}{12}=dfrac{9}{12}+dfrac{44}{12}=dfrac{53}{12}approx 4.42
$$

b.
$$
EV=12cdot 1=12
$$

a.
$$
angle E=angle Gtext{ alternate interior angles}
$$

$$
angle H=angle Ftext{ alternate interior angles}
$$

$$
Downarrow AA
$$

$$
triangle EHZsim triangle GFZ
$$

b. Corresponding sides of similar triangles are proportional:

$$
dfrac{x}{20}=dfrac{x+2}{24}
$$

Use cross multiplication:

$$
24x=20(x+2)
$$

Use distributive property:

$$
24x=20x+40
$$

Subtract $20x$ from both sides of the equation:

$$
4x=40
$$

Divide both sides of the equation by $4$:

$$
x=10
$$

a. Yes

b. $x=10$

The area of the rectangle is the product of the length and the width of the rectangle. The area is also the sum of the subareas:

a.
$$
x^2+6x+9=(x+3)(x+3)
$$

b.
$$
2x^2+5x+3=(2x+3)(x+1)
$$

c. Not possible to factor using only integers.

d.
$$
3m^2+m-14=(3m+7)(m-2)
$$

a. $(x+3)(x+3)$

b. $(2x+3)(x+1)$

c. Not favorable

c. $(3m+7)(m-2)$

The area of the rectangle is the product of the length and the width of the rectangle. The area is also the sum of the subareas:

a. We note that a term of the form $a^2-b^2$ can be factored as $(a-b)(a+b)$:

$$
9x^2=(3x-2)(3x+2)
$$

b. Factor out the greatest common factor:

$$
12x^2-16x=4x(3x-4)
$$

c.

$$
8k^2-10k+3=(4k-3)(2k-1)
$$

d. Factor out the greatest common factor:

$$
40-100m=20(2-5m)
$$

a. $(3x-2)(3x+2)$

b. $4x(3x-4)$

c. $(4k-3)(2k-1)$

d. $20(2-5m)$

The area of the rectangle is the product of the length and the width of the rectangle. The area is also the sum of the subareas:

$$
4x^2-10x-6=(x-3)(4x+2)
$$

However, you can also factor out 2 from the last factor:

$$
4x^2-10x-6=2(x-3)(x+1)
$$

Thus we note that it was possible to find more than one factorization.

b. The area of the rectangle is the product of the length and the width of the rectangle. The area is also the sum of the subareas:

$$
(x+7+3x+1)(x-2+2x+1)=x^2+5x-14+2x^2+15x+7+3x^2-5x-2+6x^2+5x+1
$$

Combine like terms:

$$
(4x+8)(3x-1)=12x^2+20x-8
$$

$$
(4x+8)(3x-1)=12x^2+20x-8
$$

The area of the rectangle is the product of the length and the width of the rectangle. The area is also the sum of the subareas:

a.
$$
k^2-12k+20=(k-10)(k-2)
$$

b.
$$
6x^2+17x-14=(3x-2)(2x+7)
$$

c.
$$
x^2-8x+16=(x-4)(x-4)=(x-4)^2
$$

d.
$$
9m^2-1=(3m-1)(3m+1)
$$

e. Each expression is quadratic because they contain the square of a variable.

a. $(k-10)(k-2)$

b. $(3x-2)(2x+7)$

c. $(x-4)^2$

d. $(3m-1)(3m+1)$

e. All are quadratic (due to the square in each expression)

The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
tan{80text{textdegree}}=dfrac{x}{58}
$$

Multiply both sides of the equation by $58$:

$$
329approx 58tan{80text{textdegree}}=x
$$

Add the eye level to the height: $329+1.5=330.5m$

The two triangles are similar because of AA. Corresponding sides of similar triangles are proportional:

$$
dfrac{x+5}{5}=dfrac{10}{4}
$$

Multiply both sides of the equation by 5:

$$
x+5=dfrac{10cdot 5}{4}=12.5
$$

Subtract 5 from both sides of the equation:

$$
x=7.5
$$

$$
x=7.5
$$

The probability is the number of favorable outcomes divided by the number of possible outcomes:

a.
$$
dfrac{14+10}{9+14+10+7}=dfrac{24}{40}=dfrac{3}{5}=0.6=60%
$$

b. Let $x$ be the total number of fish:

$$
dfrac{18}{x}=30%=0.3
$$

Multiply both sides of the equation by $x$:

$$
18=0.3x
$$

Divide both sides of the equation by 0.3:

$$
60=x
$$

Thus there are 60 fish in total.

The upper cell contains the product of the left and right cell, the lower cell contains the sum of the left and right cell.

a. Upper: $dfrac{1}{2}cdot dfrac{1}{2}=dfrac{1}{4}$ and Lower: $dfrac{1}{2}+dfrac{1}{2}=1$

b. Upper: $dfrac{1}{3}cdot dfrac{3}{4}=dfrac{1}{4}$ and Lower: $dfrac{1}{3}+dfrac{3}{4}=dfrac{4}{12}+dfrac{9}{12}=dfrac{13}{12}$

c. Upper: $xcdot x=x^2$ and Lower: $x+x=2x$

d. Upper: $acdot b=ab$ and Lower: $a+b$

a. $frac{1}{4}$, 1

b. $frac{1}{4}$, $frac{13}{12}$

c. $x^2$, $2x$

d. $ab$, $a+b$

Group like terms:

$$
(3x^3-3x^3)+(-2x^2-5x^2)+(4-(-2))
$$

Simplify:

$$
-7x^2+6
$$

a. Terms are separated by a + (or -) sign and thus the (simplified) expression contains 2 terms.

b. We note that the expression contains no $x$ term and thus the coefficient is 0.

The area of the rectangle is the product of the length and the width of the rectangle. The area is also the sum of the subareas:

a.
$$
9x^2-12x+4=(3x-2)(3x-2)=(3x-2)^2
$$

b.
$$
81m^2-1=(9m-1)(9m+1)
$$

c.
$$
x^2-11x+28=(x-7)(x-4)
$$

d.
$$
3n^2+9n+6=(3n+3)(3n+2)
$$

a. $(3x-2)^2$

b. $(9m-1)(9m+1)$

c. $(x-7)(x-4)$

d. $(3n+3)(3n+2)$

a. Yes, in the previous exercise we found the factorization:

$$
3n^2+9n+6=(3n+3)(n+2)
$$

You could also factor out the 3 in the first factor:

$$
3n^2+9n+6=3(n+1)(n+2)
$$

b. Because the expression has a greatest common factor that is not 1.

c. (i) More than one, because 2 is the greatest compmon factor

(ii) Not more than one, because 1 is the greatest common factor

(iii) More than one, because 5 is the greatest compmon factor

(iv) Not more than one, because 1 is the greatest common factor

a. Factor out the greatest common factor 5:

$$
10x^2+25x-15=5(2x^2+5x-3)
$$

b.
$$
10x^2+25x-15=5(2x^2+5x-3)=5(2x-1)(x+3)
$$

a. $5(2x^2+5x-3)$

b. $5(2x-1)(x+3)$

a. Factor out the greatest common factor 5:

$$
5x^2+15x-20=5(x^2+3x-4)
$$

Factor further using an area model:

$$
5x^2+15x-20=5(x^2+3x-4)=5(x-1)(x+4)
$$

b. Factor out the greatest common factor $3x$:

$$
3x^3-6x^2-45x=3x(x^2-2x-15)
$$

Factor further using an area model:

$$
3x^3-6x^2-45x=3x(x^2-2x-15)=3(x-5)(x+3)
$$

c. Factor out the greatest common factor 2:

$$
2x^2-50=2(x^2-25)
$$

Factor further using an area model:

$$
2x^2-50=2(x^2-25)=2(x-5)(x+5)
$$

d. Factor out the greatest common factor $y$:

$$
x^2y-3xy-10y=y(x^2-3x-10)
$$

Factor further using an area model:

$$
x^2y-3xy-10y=y(x^2-3x-10)=y(x-5)(x+2)
$$

a. $5(x-1)(x+4)$

b. $3(x-5)(x+3)$

c. $2(x-5)(x+5)$

d. $y(x-5)(x+2)$

a. The integers are closer under addition and closed under subtraction.

$$
7+(-3)=4
$$

$$
7-(-3)=10
$$

$$
4+9=13
$$

$$
4-9=-5
$$

b. Because adding two integers will be an integer too (it cannot become a rational number) and subtracting two integers will also be an integer.

c. The integers are closed under multiplication, because the product of two integers is an integer too. The integers are NOT closed under division, because the division of some integers will result in a rational number.

$$
2cdot 7=14
$$

$$
3cdot 4=12
$$

$$
4div 2=dfrac{4}{2}=2
$$

$$
4div 3=dfrac{4}{3}
$$

The area of the rectangle is the product of the length and the width of the rectangle. The area is also the sum of the subareas:

a.
$$
2x^2+3x-5=(2x+5)(x-1)
$$

b.
$$
x^2-x-6=(x-3)(x+2)
$$

c.
$$
3x^2+13x+4=(3x+1)(x+4)
$$

d. Not possible, it would however be possible it the coefficient of $x$ would be $-5$ in stead of 5.

e.
$$
7x^2-7x-42=(7x-21)(x+2)=7(x-3)(x+2)
$$

f.
$$
6x^2+26x+8=(6x+2)(x+4)=2(3x+1)(x+4)
$$

a. $(2x+5)(x-1)$

b. $(x-3)(x+2)$

c. $(3x+1)(x+4)$

d. Not factorable

e. $(7x-21)(x+2)$

f. $(6x+2)(x+4)$

a. Corresponding sides of similar triangles are proportional:

$$
dfrac{x}{10}=dfrac{30}{12}=2.5
$$

Multiply both sides of the equation by 10:

$$
x=25
$$

b. The area of a rectangle is the product of the length and the width:

$$
First: 10cdot 4=40
$$

$$
Second: 25cdot 10=250
$$

The area of a triangle isthe product of the base and the height divided by 2:

$$
First: dfrac{4cdot 8}{2}=16
$$

$$
Second: dfrac{10cdot 20}{2}=100
$$

The total area is then the sum of the subareas:

$$
First: 40+16=56
$$

$$
Second: 250+100=350
$$

a. $x=25$

b. First polygon 56 and Second polygon 350

a. The two triangles are similar because AA (corresponding angles are congruent if the lines are parallel). Corresponding sides of similar triangles are proportional:

$$
dfrac{x+13}{13}=dfrac{8+5}{8}=dfrac{13}{8}
$$

Multiply both sides of the equation by 13:

$$
x+13=dfrac{169}{8}
$$

Subtract 13 from both sides of the equation:

$$
x=dfrac{65}{8}=8.125
$$

c. The two triangles are similar because AA (alternate interior angles are congruent if the lines are parallel). Corresponding sides of similar triangles are proportional:

$$
dfrac{x}{8}=dfrac{8}{6}=dfrac{4}{3}
$$

Multiply both sides of the equation by 8:

$$
x=dfrac{32}{3}
$$

a. $x=8.125$

b. Cannot be determined

c. $x=frac{32}{3}$

a. The first dish has four choices and the second dish has three choices:

$$
4cdot 3=12
$$

b. Yes, because every main dish has an equal chance of being chosen and every dessert has an equal chance of being chosen.

c. The probability is the number of favorable outcomes divided by the total number of possible outcomes.

The probability to pick a menu without meat is then:

$$
dfrac{2}{4}cdot dfrac{3}{3}=dfrac{6}{12}=dfrac{1}{2}=0.5=50%
$$

The probability to pick a menu with chocalate is then:

$$
dfrac{4}{4}cdot dfrac{2}{3}=dfrac{2}{3}approx 0.667=66.7%
$$

a. 12 menus

b. Yes

c. $frac{1}{2}=0.5=50%$, $frac{2}{3}approx 0.667=66.7%$

If an expression is of the form $a^2-b^2$ then the factorization is $(a-b)(a+b)$.

If an expression of the form $a^2-2ab+b^2$ then the factorization is $(a-b)^2$.

If an expression of the form $a^2+2ab+b^2$ then the factorization is $(a+b)^2$.

a. $x^2-49=(x-7)(x+7)$

b. $x^2+2x-24=(x+6)(x-4)$

c. $x^2-10x+25=(x-5)(x-5)=(x-5)^2$

d. $9x^2+12x+4=(3x+2)(3x+2)=(3x+2)^2$

e. $5x^2-4x-1=(x-1)(5x+1)$

f. $4x^2-25=(2x-5)(2x+5)$

g. $x^2-6x+9=(x-3)(x-3)=(x-3)^2$

h. $x^2-36=(x-6)(x+6)$

i. $7x^2-20x-3=(7x+1)(x-3)$

j. $4x^2+20x+25=(2x+5)(2x+5)=(2x+5)^2$

k. $x^2+4$ not factorable

l. $9x^2-1=(3x-1)(3x+1)$

a. $(x-7)(x+7)$

b. $(x+6)(x-4)$

c. $(x-5)^2$

d. $(3x+2)^2$

e. $(x-1)(5x+1)$

f. $(2x-5)(2x+5)$

g. $(x-3)^2$

h. $(x-6)(x+6)$

i. $(7x+1)(x-3)$

j. $(2x+5)^2$

k. Not factorable

l. $(3x-1)(3x+1)$

If an expression is of the form $a^2-b^2$ then the factorization is $(a-b)(a+b)$.

If an expression is of the form $a^2-2ab+b^2$ then the factorization is $(a-b)^2$.

If an expression is of the form $a^2+2ab+b^2$ then the factorization is $(a+b)^2$.

If an expression is of the form $x^2+(a+b)x+(ab)$ then the factorization is $(x+a)(x+b)$.

a. $25x^2-1=(5x-1)(5x+1)$

b. $x^2-5x+36=(x-9)(x+4)$

c. $x^2+8x+16=(x+4)^2$

d. $9x^2-12x+4=(3x-2)^2$

e. This expression does not contain one of the patterns and is also not factorable.

f. $9x^2-100=(3x-10)(3x+10)$

a. $(5x-1)(5x+1)$

b. $(x-9)(x+4)$

c. $(x+4)^2$

d. $(3x-2)^2$

e. Not factorable

f. $(3x-10)(3x+10)$

The area of the rectangle is the product of the length and the width of the rectangle. The area is also the sum of the subareas:

a.
$$
u^2-x^2=(u-w)(u+w)
$$

b.
$$
u^2+2uw+w^2=(u+w)(u+w)=(u+w)^2
$$

a. $u^2-x^2=(u-w)(u+w)$

b. $u^2+2uw+w^2=(u+w)(u+w)=(u+w)^2$

a. If you add two polynomials, you obtain another polynomial and thus the set of polynomials is closed under addition:

$$
(x^2+4x+8)+(4x^2+3x+7)=5x^2+7x+15
$$

$$
(4x+8)+(3x+7)=7x+15
$$

If you subtract two polynomials, you obtain another polynomial and thus the set of polynomials is closed under subtraction:

$$
(x^2+4x+8)-(4x^2+3x+7)=-3x^2+x+1
$$

$$
(4x+8)+(3x+7)=x+1
$$

b. Because the sum and difference of polynomial will always remain of the same form and will thus remain a polynomial.

If an expression is of the form $a^2-b^2$ then the factorization is $(a-b)(a+b)$.

If an expression is of the form $a^2-2ab+b^2$ then the factorization is $(a-b)^2$.

If an expression is of the form $a^2+2ab+b^2$ then the factorization is $(a+b)^2$.

If an expression is of the form $x^2+(a+b)x+(ab)$ then the factorization is $(x+a)(x+b)$.

**Concept**
The special products have the following forms:
$$begin{gather}
(a^2-b^2) = (a+b)(a-b) \
(a^2+2ab+b^2) = (a+b)^2 \
(a^2-2ab+b^2) = (a+b)^2
end{gather}$$

**Part a**
We factor the polynomial $x^2-64$ using Eq (1) as follows:
$$begin{aligned}
(a^2-b^2) &= (a+b)(a-b) \
x^2 – 64 &= (x+b)(x-b) \
x^2 – 64 &= (x+8)(x-8) \
end{aligned}$$

Hence, the factors $(x+8)(x-8)$ yields the polynomial $x^2-64$

**Part b**
We factor the polynomial $x^2-6x+9$ using Eq (3) as follows:
$$begin{aligned}
(a^2-2ab+b^2) &= (a-b)^2 \
x^2 – 6x+9 &= (x-b)^2 \
x^2 – 6x+9 &= (x-3)^2 \
end{aligned}$$

Hence, the factors $(x-3)(x-3)$ yields the polynomial $x^2 – 6x+94$

**Part c**
We factor the polynomial $4x^2+4x+1$ using Eq (2) as follows:
$$begin{aligned}
(a^2+2ab+b^2) &= (a+b)^2 \
4x+4x+1 &= (2x+b)^2 \
4x+4x+1 &= (2x+1)^2 \
end{aligned}$$

Hence, the factors $(2x+1)(2x+1)$ yields the polynomial $4x+4x+1$

**Part d**
We factor the polynomial $4x^2-49$ using Eq (1) as follows:
$$begin{aligned}
(a^2-b^2) &= (a+b)(a-b) \
4x^2 – 49 &= (2x+b)(2x-b) \
4x^2 – 49 &= (2x+7)(2x-7) \
end{aligned}$$

Hence, the factors $(2x+7)(2x-7)$ yields the polynomial $4x^2 – 49$

a. $(x-8)(x+8)$
b. $(x-3)(x-3)$
c. $(2x+1)(2x+1)$
d. $(2x-7)(2x+7)$

Prime number is a number greater than one that is not a product of two smaller numbers. It is only divisible by $1$ and the number itself.

Let $W=$ set of prime numbers.
$$begin{aligned}
W=2,3,5,7,11,13,17,….
end{aligned}$$

Let’s add the prime numbers.
$$begin{aligned}
2+3&=5enspacetext{textcolor{#4257b2}{(prime number)}}\
2+5&=7enspacetext{textcolor{#4257b2}{(prime number)}}\
2+7&=9enspacetext{textcolor{#4257b2}{(not a prime number)}}\
end{aligned}$$

Based on the result in **Step 5**, there is one answer that is not a prime number. Therefore, **prime numbers** are $text{textcolor{#4257b2}{not}}$ **closed set under addition**.

**Concept**
The Pythagorean theorem can be used to determine the relationship between the sides of the triangle. This can be described by the expression:
$$begin{gather}
c^2 = a^2 + b^2
end{gather}$$

**Part I: Area**
We determine the area of the figure by separating the figure into two parts: triangle and rectangle. We first calculate for the area of the rectangle as follows:
$$begin{aligned}
A_{text{square}} &= text{length} times text{width} \
&= 16.0 times 18.0 \
&= 288 text{ m}^2
end{aligned}$$

Afterwards we determine the area of the triangular part of the figure. The formula for getting the area of the triangle is described by the expression: $A_{text{triangle}} = frac{1}{2}bh$. However, we do not have the values for the base and height of the triangle. We can obtain the height of the triangle by subtracting the height difference in the figure as follows:
$$begin{aligned}
h &= 24.0 – 18.0 \
h &= 6.0
end{aligned}$$

From the figure, the opposite side with respect to the angle is the base of the triangle. We know that the trigonometric identity $tan theta = frac{text{opposite}}{text{adjacent}}$ can be used to determine the base of the triangular part. Using this expression, we obtain the base as follows:
$$begin{aligned}
tan theta &= frac{text{opposite}}{text{adjacent}} \
text{opposite} &= text{adjacent} cdot tan theta \
b &= h cdot tan theta \
b &= (6.0) cdot tan (20^circ) \
b &= 2.18 text{ m}
end{aligned}$$

Now that we have identified the values of the height and base of the triangular part, we obtain the area of the triangle as follows:
$$begin{aligned}
A_{text{triangle}} &= frac{1}{2}bh \
&= frac{1}{2} (2.18) cdot (6.0) \
&= boxed{6.54 text{ m}^2}
end{aligned}$$

To be able to calculate the area of the figure, we add the area of the square and triangular parts of the figure as follows:
$$begin{aligned}
A_{text{total}} &= A_{text{square}} + A_{text{triangle}} \
&= 288+ 6.54 \
&= boxed{294.54 text{ m}^2}
end{aligned}$$

Since we were able to obtain the base of the triangular part, we use this value and the length of the square to determine the length of the top portion of the figure as follows:
$$begin{aligned}
l &= 16.0 – b_{text{triangle}} \
&= 16.0 – 2.8 \
&= 13.2 text{ m}
end{aligned}$$

We consider that the base and height of the triangle can be used to determine the hypotenuse of the triangle from the Pythagorean theorem. We determine the value of the remaining side of the triangle (hypotenuse) using Eq (1) as follows:
$$begin{aligned}
c^2 &= a^2 + b^2 \
c &= sqrt{a^2 + b^2} \
c &= sqrt{b^2 + h^2} \
c &= sqrt{(2.8)^2 + (6.0)^2} \
c &=6.62 text{ m}
end{aligned}$$

Hence, the length of the remaining side is $6.62 text{ m}$

Now that we have obtained all the values of the sides of the figure, we add these values to determine the perimeter of the figure as follows:
$$begin{aligned}
P &= 24.0 + 16.0 + 18.0 + 6.62 + 13.2 \
& boxed{P = 77.82 text{ m}}
end{aligned}$$

The perimeter of the figure is $77.82 text{ m}$

$A= 294.54 text{ m}^2$
$P = 77.82 text{ m}$

c. The graph of $y=-x^2$ is the the graph of $y=x^2$ reflected about the $x$-axis.

d. On the graph we note $xapprox pm 2.24$.

e. On the graph we note $xapprox pm 3.16$.

The graph of $y=-x^2$ is the the graph of $y=x^2$ reflected about the $x$-axis.

a. The intersection counts the possibilities that the sum is a multiple of 3 and is a multiple of 4. Thus the intersection counts the possibilities that the sum is a multiple of 12 and thus contains only 1 outcome: 12(=6+6) itself.

b. Use the addition rule:

$$
P(Atext{ or B})=P(A)+P(B)-P(Atext{ and }B)=dfrac{12}{36}+dfrac{9}{36}-dfrac{1}{36}=dfrac{20}{36}=dfrac{5}{9}approx 0.555=55.5%
$$

No, because both $Delta y$ and $Delta x$ have not been given and thus we have two unknown values for one known equation, which is not enough to determine the value of the unknown values.

a. METHOD 1: Determine $Delta y$ using the Pythagorean theorem:

$$
Delta y=sqrt{6^2-3^2}=sqrt{27}=3sqrt{3}approx 5.2
$$

METHOD 2: The tangent ratio is the opposite side divided by the afjacent rectangular side:

$$
tan{60text{textdegree}}=dfrac{Delta y}{3}
$$

Multiply both sides of the equation by 3:

$$
5.2approx 3tan{60text{textdegree}}=Delta y
$$

b. The ratios must be equal, because the triangle are similar (SAS).
$$
dfrac{Delta x}{hypotenuse}=dfrac{1}{2}=dfrac{3}{6}=dfrac{1}{2}
$$

c. The ratio of $Delta x$ to the hypotenuse should be equal to $dfrac{1}{2}$:

$$
dfrac{BC}{7}=dfrac{1}{2}
$$

Multiply both sides of the equation by 7:

$$
BC=dfrac{7}{2}
$$

Determine $AC$ using the Pythagorean theorem:

$$
AC=sqrt{7^2-left( dfrac{7}{2}right)^2}=sqrt{dfrac{147}{4}}=dfrac{sqrt{147}}{2}
$$

a. $Delta y=5.2$

b. $frac{1}{2}$

c. $BC=frac{7}{2}$ and $AC=frac{sqrt{147}}{2}$

a. If the sine/cosine is a rational number, then the calculator will give the correct result. If the sine/cosine is not a rational number, then the calculator will give a numerical approximation.

$$
sin{60text{textdegree}}approx 0.866
$$

$$
cos{60text{textdegree}}=0.5
$$

b. The sine ratio is the opposite side divided by the hypotenuse:

$$
sin{23text{textdegree}}=dfrac{a}{15}
$$

Multiply both sides of the equation by 15:

$$
5.86approx 15sin{23text{textdegree}}=a
$$

c. The cosine ratio is the adjacent rectangular side divided by the hypotenuse:

$$
cos{37text{textdegree}}=dfrac{b}{8}
$$

Multiply both sides of the equation by 8:

$$
4.81approx 8cos{37text{textdegree}}=b
$$

a. Yes

b. Sine ratio, $a=5.86$

b. $b=4.81$

The sine ratio is the opposite side divided by the hypotenuse:

$$
sin{31.5text{textdegree}}=dfrac{Delta y}{100}
$$

Multiply both sides of the equation by 100:

$$
52.25approx 100sin{31.5text{textdegree}}=Delta y
$$

The cosine ratio is the adjacent rectangular side divided by the hypotenuse:

$$
cos{31.5text{textdegree}}=dfrac{Delta x}{100}
$$

Multiply both sides of the equation by 100:

$$
85.26approx 100cos{31.5text{textdegree}}=Delta x
$$

$$
Delta yapprox 52.25ft
$$

$$
Delta xapprox 85.26ft
$$

a. Adjacent rectangular side has been given.The cosine ratio is the adjacent rectangular side divided by the hypotenuse:

$$
cos{17text{textdegree}}=dfrac{3}{x}
$$

Multiply both sides of the equation by $x$:

$$
xcos{17text{textdegree}}=3
$$

Divide both sides of the equation by $cos{17text{textdegree}}$:

$$
x=dfrac{3}{cos{17text{textdegree}}}approx 3.14
$$

b. Opposite side has been given. The sine ratio is the opposite side divided by the hypotenuse:

$$
sin{25text{textdegree}}=dfrac{x}{9}
$$

Multiply both sides of the equation by 9:

$$
3.8approx 9sin{25text{textdegree}}=x
$$

c. Adjacent rectangular side has been given. The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
tan{45text{textdegree}}=dfrac{x}{5}
$$

Multiply both sides of the equation by 5:

$$
5=5tan{45text{textdegree}}=x
$$

d. Adjacent rectangular side has been given.The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
tan{62text{textdegree}}=dfrac{5}{x}
$$

Multiply both sides of the equation by $x$:

$$
xtan{62text{textdegree}}=5
$$

Divide both sides of the equation by $tan{62text{textdegree}}$:

$$
x=dfrac{5}{tan{62text{textdegree}}}approx 2.66
$$

e. Opposite side has been given. The sine ratio is the opposite side divided by the hypotenuse:

$$
sin{34text{textdegree}}=dfrac{13}{x}
$$

Multiply both sides of the equation by $x$:

$$
xsin{34text{textdegree}}=13
$$

Divide both sides of the equation by $sin{34text{textdegree}}$:

$$
x=dfrac{13}{sin{34text{textdegree}}}approx 23.25
$$

f. Adjacent rectangular side has been given. The cosine ratio is the adjacent rectangular side divided by the hypotenuse:

$$
cos{20text{textdegree}}=dfrac{10}{x}
$$

Multiply both sides of the equation by $x$:

$$
xcos{20text{textdegree}}=10
$$

Divide both sides of the equation by $cos{20text{textdegree}}$:

$$
x=dfrac{10}{cos{20text{textdegree}}}approx 10.64
$$

a. $x=3.14$

b. $x=3.8$

c. $x=5$

d. $x=2.66$

e. $x=23.25$

f. $x=10.64$

The cosine ratio is the adjacent rectangular side divided by the hypotenuse.

The sine ratio is the opposite side divided by the hypotenuse.

The tangent ratio is the opposite side divided by the adjacent rectangular side.

The Pythagorean theorem states that the sum of the squares of the legs is equal to the square of the hypotenuse:

$$
a^2+b^2=c^2
$$

The sine ratio is the opposite side divided by the hypotenuse:

$$
sin{47text{textdegree}}=dfrac{y}{11}
$$

Multiply both sides of the equation by 11:

$$
8.04approx 11sin{47text{textdegree}}=y
$$

The cosine ratio is the adjacent rectangular side divided by the hypotenuse:

$$
cos{47text{textdegree}}=dfrac{x}{11}
$$

Multiply both sides of the equation by 11:

$$
7.5approx 11cos{47text{textdegree}}=x
$$

$$
yapprox 8.04
$$

$$
xapprox 7.5
$$

a. The sine ratio is the opposite side divided by the hypotenuse.

$$
sin{theta}=dfrac{b}{a}
$$

b. The tangent ratio is the opposite side divided by the adjacent rectangular side.

$$
tan{theta}=dfrac{a}{b}
$$

c. The cosine ratio is the adjacent rectangular side divided by the hypotenuse.

$$
cos{theta}=dfrac{a}{b}
$$

a. $sin{theta}=dfrac{b}{a}$

b. $tan{theta}=dfrac{a}{b}$

c. $cos{theta}=dfrac{a}{b}$

Factorize the polynomials:

a. $6x^2+5x-6=(3x-2)(2x+3)$

b. $8x^2-50=2(4x^2-25)=2(2x-5)(2x+5)$

c. $2x^3+2x^2-112x=2x(x^2+x-56)=2x(x+8)(x-7)$

d. $9x^2-24x+16=(3x-4)(3x-4)=(3x-4)^2$

a. $(3x-2)(2x+3)$

b. $2(2x-5)(2x+5)$

c. $2x(x+8)(x-7)$

d. $(3x-4)^2$

a. The sides opposite the equal angles, need to have equal length:

$$
ED=EF
$$

b. Let us first determine the height of the triangle. We note that the height is the opposite side of the angle 54$text{textdegree}$ in a right triangle. The sine ratio is the length of the opposite side divided by the length of the hypotenuse in a right triangle.

$$
sin(54text{textdegree})=frac{text{Height}}{DE}
$$

Solve the equation to the height:

$$
begin{align*}
text{Height}&=DEcdot sin (54text{textdegree})=9sin (54text{textdegree})approx 7.2812
end{align*}
$$

Next, we determine $x$ using the Pythagorean theorem $a^2+b^2=c^2$ (where $x$ is half the length of $overline{DF}$).

$$
begin{align*}
x&=sqrt{ED^2-text{Height}^2}=sqrt{9^2-7.2812^2}approx 5.2900
end{align*}
$$

Finally, the length of $overline{DF}$ is twice $x$:

$$
DF=2x=2(5.2900)=10.5800
$$

a. $ED=EF$

b. $DF=10.5800$ mm

b. The probability is the number of favorable outcomes divided by the number of possible outcomes:

(i)
$$
P(odd)=dfrac{3}{4}=0.75=75%
$$

(ii)
$$
P(text{Not letter})=dfrac{2}{3}approx 0.667=66.7%
$$

(iii)
$$
P(28text{ and }book)=dfrac{1}{4}cdot dfrac{1}{3}=dfrac{1}{12}approx 0.083=8.3%
$$

(iv)
$$
P(text{Not read})=dfrac{2}{3}approx 0.667=66.7%
$$

c. No, because the activities were randomly chosen.

a. Tree diagram

b.

(i) $frac{3}{4}=0.75=75%$

(ii) $frac{2}{3}approx 0.667=66.7%$

(iii) $frac{1}{12}approx 0.083=8.3%$

(iv) $frac{2}{3}approx 0.667=66.7%$

c. No

b. The probability is the number of favorable outcomes divided by the number of possible outcomes:

$$
dfrac{2}{4}=frac{1}{2}=0.5=50%
$$

a. Tree diagram

b. $frac{1}{2}=0.5=50%$

a. The sine ratio is the opposite side divided by the hypotenuse:

$$
sin{28text{textdegree}}=dfrac{x}{19}
$$

Multiply both sides of the equation by 19:

$$
8.92approx 19sin{28text{textdegree}}=x
$$

b. The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
tan{45text{textdegree}}=dfrac{8}{x}
$$

Multiply both sides of the equation by $x$:

$$
xtan{45text{textdegree}}=8
$$

Divide both sides of the equation by $tan{45text{textdegree}}$

$$
x=dfrac{8}{tan{45text{textdegree}}}=8
$$

c. The cosine ratio is the adjacent rectangular side divided by the hypotenuse:

$$
cos{61text{textdegree}}=dfrac{18}{x}
$$

Multiply both sides of the equation by $x$:

$$
xcos{61text{textdegree}}=18
$$

Divide both sides of the equation by $cos{61text{textdegree}}$:

$$
x=dfrac{18}{cos{61text{textdegree}}}approx 37.13
$$

d. The sine ratio is the opposite side divided by the hypotenuse:

$$
sin{25text{textdegree}}=dfrac{6}{x}
$$

Multiply both sides of the equation by $x$:

$$
xsin{25text{textdegree}}=6
$$

Divide both sides of the equation by $sin{34text{textdegree}}$:

$$
x=dfrac{6}{sin{25text{textdegree}}}approx 14.2
$$

e. The cosine ratio is the adjacent rectangular side divided by the hypotenuse:

$$
cos{60text{textdegree}}=dfrac{8}{x}
$$

Multiply both sides of the equation by $x$:

$$
xcos{60text{textdegree}}=8
$$

Divide both sides of the equation by $cos{60text{textdegree}}$:

$$
x=dfrac{8}{cos{60text{textdegree}}}=16
$$

f. The sine ratio is the opposite side divided by the hypotenuse:

$$
sin{30text{textdegree}}=dfrac{x}{42}
$$

Multiply both sides of the equation by 42:

$$
21=42sin{30text{textdegree}}=x
$$

g. The triangle does not exist, because an angle of a triangle cannot be zero.

h. The cosine ratio is the adjacent rectangular side divided by the hypotenuse:

$$
cos{73text{textdegree}}=dfrac{x}{3}
$$

Multiply both sides of the equation by $3$:

$$
0.88approx 3cos{73text{textdegree}}=x
$$

i. The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
tan{42text{textdegree}}=dfrac{x}{12}
$$

Multiply both sides of the equation by $12$:

$$
10.8approx 12tan{42text{textdegree}}=x
$$

a. $x=8.92$

b. $x=8$

c. $x=37.13$

d. $x=14.2$

e. $x=16$

f. $x=21$

g. Triangle doesn’t exist

h. $x=0.88$

i. $x=10.8$

a. The sine ratio is the opposite side divided by the hypotenuse. Next, multiply both sides of the equation by $x$. Then divide both sides of the equation by $sin{25text{textdegree}}$.

b.
$$
x=dfrac{29}{sin{25text{textdegree}}}approx 68.62
$$

c. Because Ziv used the other angle in the triangle and the cosine is the adjacent rectangular side divided by the hypotenuse.

d. METHOD 1: The sine ratio is the opposite side divided by the hypotenuse.

$$
sin{53text{textdegree}}=dfrac{y}{5}
$$

Multiply both sides of the equation by 5:

$$
3.99approx 5sin{53text{textdegree}}=y
$$

METHOD 2: The sum of all angles in a triangle is 180$text{textdegree}$:

$$
180text{textdegree}-53text{textdegree}-90text{textdegree}=37text{textdegree}
$$

The cosine ratio is the adjacent rectangular side divided by the hypotenuse.

$$
cos{37text{textdegree}}=dfrac{y}{5}
$$

Multiply both sides of the equation by 5:

$$
3.99approx 5cos{37text{textdegree}}=y
$$

a. The sine ratio is the opposite side divided by the hypotenuse. Next multiply both sides of the equation by $x$. Then divide both sides of the equation by $sin{25text{textdegree}}$.

b. 68.62

c. Because Ziv used the other angle in the triangle and the cosine is the adjacent rectangular side divided by the hypotenuse.

d. $y=3.99$

a. The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
tan{theta}=dfrac{9}{9}=1
$$

We then know that $theta=45text{textdegree}$.

b. The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
tan{theta}=dfrac{1}{5}
$$

We then know that $theta=11text{textdegree}$.

c. The cosine ratio is the opposite side divided by the adjacent rectangular side:

$$
cos{theta}=dfrac{5}{10}=dfrac{1}{2}
$$

We then know that $theta=60text{textdegree}$.

a. $theta=45text{textdegree}$

b. $theta=11text{textdegree}$

c. $theta=60text{textdegree}$

a. The sine ratio is the opposite side divided by the hypotenuse:

$$
sin{22text{textdegree}}=dfrac{x}{17}
$$

Multiply both sides of the equation by 17:

$$
6.37approx 17sin{22text{textdegree}}=x
$$

b. The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
tan{49text{textdegree}}=dfrac{7}{x}
$$

Multiply both sides of the equation by $x$:

$$
xtan{49text{textdegree}}=7
$$

Divide both sides of the equation by $tan{49text{textdegree}}$:

$$
x=dfrac{7}{tan{49text{textdegree}}}
$$

c. The cosine ratio is the adjacent rectangular side divided by the hypotenuse:

$$
cos{60text{textdegree}}=dfrac{x}{6}
$$

Multiply both sides of the equation by $6$:

$$
3=6cos{60text{textdegree}}=x
$$

d. The sum of all angles in a triangle is 180$text{textdegree}$:

$$
180text{textdegree}-90text{textdegree}-60text{textdegree}=30text{textdegree}
$$

The sine ratio is the opposite side divided by the hypotenuse:

$$
sin{30text{textdegree}}=dfrac{x}{6}
$$

Multiply both sides of the equation by $6$:

$$
3=6sin{30text{textdegree}}=x
$$

a. $x=6.37$

b. $x=frac{7}{tan 49text{textdegree}}$

c. $x=3$

d. $x=3$

a. The cosine ratio is the adjacent rectangular side divided by the hypotenuse and since the angle of 29$text{textdegree}$ is adjacent to the side with the unknown length $y$ we should use this angle.

b.
$$
cos{29text{textdegree}}=dfrac{y}{42}
$$

Multiply both sides of the equation by $42$:

$$
36.73approx 42cos{29text{textdegree}}=y
$$

a. 29$text{textdegree}$

b. $y=36.73$

a. Corresponding sides of similar triangles are proportional:

$$
dfrac{m}{22}=dfrac{15}{10}
$$

Multiply both sides of the equation by 22:

$$
m=dfrac{15cdot 22}{10}=33
$$

Determine the other side:

$$
dfrac{n}{24}=dfrac{15}{10}
$$

Multiply both sides of the equation by 24:

$$
n=dfrac{15cdot 24}{10}=36
$$

b. The area of a rectangle is the product of the length and the width. The area of a triangle is the product of the base and the height divided by 2. Determine then the area of both figure by determining the area of the rectangle and decrease the area by the area of the missing triangles:

$$
FIRST: 22cdot 24-dfrac{24cdot 10}{2}-dfrac{5cdot 12}{2}=528-120-30=378
$$

$$
SECOND: 378cdot left(dfrac{15}{10}right)^2=850.5
$$

The perimeter is the sum of all sides (determine the missing side lengths using the Pythagorean theorem):

$$
FIRST=22+19+13+26=80
$$

$$
SECOND=80cdot dfrac{15}{10}=120
$$

a. $m=33$ and $n=36$

b. First figure: Area 378 and perimeter 80

Second figure: Area 850.5 and perimeter 120

The area of the rectangle is the product of the length and the width of the rectangle. The area is also the sum of the subareas:

a.
$$
xy-6x-18+3y=(x+3)(y-6)
$$

b.
$$
2xy-3y+2x-3=(2x-3)(y+1)
$$

c.
$$
12y+14x-8xy-21=(3-2x)(4y-7)
$$

d. Place terms that both contain an $x$ term in the same row (or column) and the same for terms that contain both a $y$ term, the remaining term then is place in the remaining cell.

a. $(x+3)(y-6)$

b. $(2x-3)(y+1)$

c. $(3-2x)(4y-7)$

d. lace terms that both contain an $x$ term in the same row (or column) and the same for terms that contain both a $y$ term, the remaining term then is place in the remaining cell.

The graph is a parabola with vertex $(-1,-4)$ and the parabola opens upwards. The $x$-intercept is $x=-3$ and $x=1$, while the $y$-intercept is $y=-3$. The function is positive on $x1$, while the function is negative on $-3<x<1$.

$x$-intercepts $(-3,0)$ and $(1,0)$

$y$-intercept $(0,-3)$

The expected value is the sum of the products of the possibilities and their probability:

a.
$$
EV=4cdot dfrac{1}{2}-7cdot dfrac{1}{2}=dfrac{4}{2}-dfrac{7}{2}=-dfrac{3}{2}=-1.5
$$

b. The expected winings (or losings) is the product of the expected value andthe number of flips:

$$
8cdot left( -dfrac{3}{2}right)=-12
$$

a. -$1.5

b. Loss of $12

b.
$$
cos^{-1}{dfrac{8}{16}}=60text{textdegree}
$$

c. The tangent is the opposite side divided by the adjacent rectangular side:

$$
x=tan^{-1}left( dfrac{3}{25}right)=6.84text{textdegree}
$$

Thus we note that the ramp will not meet the state building code.

d. The tangent is the opposite side divided by the adjacent rectangular side:

$$
tan{4.76text{textdegree}}=dfrac{3}{x}
$$

Multiply both sides of the equation by $x$:

$$
xtan{4.76text{textdegree}}=3
$$

Divide both sides of the equation by $tan{4.76text{textdegree}}$:

$$
x=dfrac{3}{tan{4.76text{textdegree}}}approx 36
$$

Thus the ramp should start about 36 ft from the building.

a. Angle

b. Yes

c. 6.84$text{textdegree}$

d. 36 ft

The sine is the opposite side divided by the hypotenuse:

$$
mangle B =sin^{-1}left( dfrac{14}{15}right)approx 69text{textdegree}
$$

The sum of all angles in a triangle is 180$text{textdegree}$:

$$
mangle A=180text{textdegree}-69text{textdegree}-90text{textdegree}=21text{textdegree}
$$

$$
mangle B =69text{textdegree}
$$

$$
mangle A=21text{textdegree}
$$

a. The cosine ratio is the adjacent rectangular side divided by the hypotenuse:

$$
theta=cos^{-1}{dfrac{12}{24}}=cos^{-1}{dfrac{1}{2}}=60text{textdegree}
$$

b. The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
theta=tan^{-1}{dfrac{8}{8}}=tan^{-1}{1}=45text{textdegree}
$$

c. The sine ratio is the opposite side divided by the hypotenuse:

$$
sin{57text{textdegree}}=dfrac{7}{x}
$$

Multiply both sides of the equation by $x$:

$$
xsin{57text{textdegree}}=7
$$

Divide both sides of the equation by $sin{57text{textdegree}}$:

$$
x=dfrac{7}{sin{57text{textdegree}}}approx 8.35
$$

a. $theta=60text{textdegree}$

b. $theta=45text{textdegree}$

c. $x=8.35$

a. If one angle of a right triangle has the tangent ratio $dfrac{a}{b}$, then the complementary angle has a tangent ratio of $dfrac{b}{a}$.

b. If one angle of a right triangle has the sine ratio $dfrac{a}{b}$, then the complementary angle has a cosine ratio of $dfrac{a}{b}$.

If one angle of a right triangle has the cosine ratio $dfrac{a}{b}$, then the complementary angle has a sine ratio of $dfrac{a}{b}$.

If one angle of a right triangle has the tangent ratio $dfrac{a}{b}$, then the complementary angle has a tangent ratio of $dfrac{b}{a}$.

METHOD 1: The cosine ratio is the adjacent rectangular side divided by the hypotenuse:

$$
angle A=cos^{-1}{dfrac{5}{13}}approx 67text{textdegree}
$$

METHOD 2: The sine ratio is the opposite side divided by the hypotenuse:

$$
angle A=sin^{-1}{dfrac{12}{13}}approx 67text{textdegree}
$$

About 67 $text{textdegree}$

If an expression is of the form $a^2-b^2$ then the factorization is $(a-b)(a+b)$.

If an expression is of the form $a^2-2ab+b^2$ then the factorization is $(a-b)^2$.

If an expression is of the form $a^2+2ab+b^2$ then the factorization is $(a+b)^2$.

If an expression is of the form $x^2+(a+b)x+(ab)$ then the factorization is $(x+a)(x+b)$.

a. Difference of squares: $4x^2-49=(2x)^2-7^2=(2x-7)(2x+7)$

b. $2x^2+11x+12=(2x+3)(x+4)$

c. Not possible

d. Perfect square trinomial: $9x^2+6x+1=(3x)^2+2(3x)(1)+1^2=(3x+1)^2$

e. $xy-4y+2x-8=y(x-4)+2(x-4)=(y+2)(x-4)$

f. Difference of squares: $x^4-16=(x^2-4)(x^2+4)=(x-2)(x+2)(x^2+4)$

a. $(2x-7)(2x+7)$

b. $(2x+3)(x+4)$

c. Not possible

d. $(3x+1)^2$

e. $(y+2)(x-4)$

f. $(x-2)(x+2)(x^2+4)$

a. Multiply both sides of the equation by 5:

$$
7-y=dfrac{15}{4}
$$

Subtract 7 from both sides of the equation:

$$
-y=-dfrac{13}{4}
$$

Multiply both sides of the equation by $-1$:

$$
y=dfrac{13}{4}
$$

b. Use cross multiplication:

$$
3(y-2)=6y
$$

Use distributive property:

$$
3y-6=6y
$$

Subtract $3y$ from both sides of the equation:

$$
-6=3y
$$

Divide both sides of the equation by $3$:

$$
-2=y
$$

c. Thus the grow ratio (per month) is
$$
dfrac{1dfrac{3}{4}}{4dfrac{1}{2}}=dfrac{1.75}{4.5}=dfrac{7}{18}
$$

The growth per year is then the product of the number of months and the grow ratio:

$$
12cdot dfrac{7}{18}=dfrac{84}{18}=dfrac{14}{3}=4dfrac{2}{3}
$$

d. Corresponding sides of similar triangles are proportional:

$$
dfrac{x}{4}=dfrac{x+8}{12}
$$

Use cross multiplication:

$$
12x=4(x+8)
$$

Use distributive property:

$$
12x=4x+32
$$

Subtract $4x$ from both sides of the equation:

$$
8x=32
$$

Divide both sides of the equation by $8$:

$$
x=4
$$

a. $y=frac{13}{4}$

b. $y=-2$

c. $4frac{2}{3}$

d. $x=4$

Addition rule: $P(Atext{ or }B)=P(A)+P(B)-P(Atext{ and }B)$:

$$
P(blemishwalktext{ or }dent)=P(blemish)+P(dent)-P(blemishtext{ and }dent)=4%+dfrac{1}{2}%-dfrac{1}{2}%=4%
$$

Thus the probability that the refrigerator has a defect is 4%, and thus 4 in 100 refrigerators will have a defect, which is reasonably small.

**Concept**
The Pythagorean theorem can be used to determine the relationship between the sides of the triangle. This can be described by the expression:
$$begin{gather}
c^2 = a^2 + b^2
end{gather}$$

**Part a**
The length of the rope between David and the carabiner above Emily is equal to the value of $x$ in the figure. Using the trigonometric identity: $cos theta = frac{text{adjacent}}{text{hypotenuse}}$, the hypotenuse side of the triangle represents $x$ while the adjacent side is the distance of David from the wall. We calculate for the value of $x$ as follows:
$$begin{aligned}
cos theta &= frac{text{adjacent}}{text{hypotenuse}} \
cos (55^circ) &= frac{20}{x} \
x &= frac{20}{cos (55^circ)} \
& boxed{x = 34.87 text{ ft}}
end{aligned}$$

Therefore, the length of the rope between David and carabiner above Emily is $34.87 text{ ft}$

**Part b**
We take note that the total length of the rope is $48 text{ ft}$. We determine the length of the rope between Emily and the carabiner above her by subtracting the total length of the rope by $x$ as follows:
$$begin{aligned}
y &= l_text{rope} – x \
y &= 48 – 34.87 \
& boxed{y = 13.13 text{ ft}}
end{aligned}$$

We use this value as the distance between Emily and the highest point of the rope (location of the carabiner).

Next, we determine the value of $b$ using the trigonometric identity: $tan theta = frac{text{opposite}}{text{adjacent}}$. We determine the value of $b$ as follows:
$$begin{aligned}
tan theta &= frac{text{opposite}}{text{adjacent}} \
tan theta &= frac{b}{20.0} \
b &= 20.0 cdot tan theta \
b &= (20.0) cdot tan (55^circ) \
& boxed{b = 28.56 text{ ft}}
end{aligned}$$

Since the rope is attached to David’s waist $3$ ft from the ground, the height of the wall can be calculated by adding the height of the rope from the ground and the value of $b$ as follows:
$$begin{aligned}
h_{text{wall}} &= b + 3 \
h_{text{wall}} &= 28.56 + 3 \
& boxed{h_{text{wall}} = 31.56 text{ ft}}
end{aligned}$$

We can determine how high Emily has climbed by deducting the height of the wall to the distance between Emily and the carabiner as follows:
$$begin{aligned}
h_{text{Emily}} &= h_{text{wall}} – y \
&= 31.56 – 13.13 \
&= boxed{18.43 text{ ft}}
end{aligned}$$

Therefore, Emily has climbed $18.43 text{ ft}$ above the ground.

$x = 34.87 text{ ft}$
$h_{text{Emily}} = 18.43 text{ ft}$

a. The sine ratio is the opposite side divided by the hypotenuse:

$$
sin{60text{textdegree}}=dfrac{70}{x}
$$

Multiply both sides of the equation by $x$:

$$
xsin{60text{textdegree}}=70
$$

Divide both sides of the equation by $sin{60text{textdegree}}$:

$$
x=dfrac{70}{sin{60text{textdegree}}}approx 81ft
$$

b. The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
tan{60text{textdegree}}=dfrac{70}{x}
$$

Multiply both sides of the equation by $x$:

$$
xtan{60text{textdegree}}=70
$$

Divide both sides of the equation by $tan{60text{textdegree}}$:

$$
x=dfrac{70}{tan{60text{textdegree}}}approx 40ft
$$

a. The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
tan{10text{textdegree}}=dfrac{x}{185}
$$

Multiply both sides of the equation by $185$:

$$
33approx 185tan{10text{textdegree}}=x
$$

The height of Emily is then $33+6=39$ ft (add the height of Nathan’s eyes).

b. The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
tan{25text{textdegree}}=dfrac{x}{185}
$$

Multiply both sides of the equation by $185$:

$$
86approx 185tan{25text{textdegree}}=x
$$

The height of Emily is then $86+6=91$ ft (add the height of Nathan’s eyes).

The distance climbed is then the difference with the previous height: 91ft$-$39ft=52ft.

c. The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
theta=tan^{-1}{dfrac{91+32}{185}}approx 34text{textdegree}
$$

a. About 39 ft

b. About 91 ft

c. About 34$text{textdegree}$

a. The sine ratio is the opposite side divided by the hypotenuse:

$$
sin{36text{textdegree}}=dfrac{x}{9}
$$

Multiply both sides of the equation by $9$:

$$
5.3approx 9sin{36text{textdegree}}=x
$$

The cosine ratio is the adjacent rectangular side divided by the hypotenuse:

$$
cos{36text{textdegree}}=dfrac{l}{9}
$$

Multiply both sides of the equation by $9$:

$$
7.3approx 9cos{36text{textdegree}}=x
$$

The area of a trapezium is then the product of the bases multiplied by the height, divided by 2:

$$
AREA=dfrac{(8+8+5.3)7.3}{2}=77.745
$$

Thus the area is 77.745 square meter $approx 837$ square ft.

b.The number of cans needed is then the area of the surface divided by the area covered per paint can, and then the result rounded up to the nearest integer:

$$
dfrac{837}{150}=5.58approx 6
$$

Thus he will need 6 cans.

Given Values.
$$begin{aligned}
d_1&=18 ftenspacetext{(distance from David to wall base)}\
d_2&=185 ftenspacetext{(distance from Nathan to wall base)}\
end{aligned}$$

Let
$$begin{aligned}
theta_1&=text{angle from David waist to the ground}\
theta_2&=text{angle of sight of Nathan to Emily}\
y&=text{height from carabiner to the ground}\
y_1&=text{height from Emily to David’s waist}\
y_2&=text{height from Emily to Nathan’s eyes}\
h&=text{distance of rope from David’s waist to Emily}\
h_1&=text{distance from Nathan’s eyes to Emily}\
end{aligned}$$

The goal is to find the value of $theta_2$. By using the Trigonometric Ratios.
$$begin{aligned}
tantheta_2&=dfrac{y_2}{d_2}\
tantheta_2&=dfrac{y_2}{185}
end{aligned}$$

Solve for the value of $theta_1$.
$$begin{aligned}
theta_1&=180degree-90degree-6degree\
theta_1&=84degree
end{aligned}$$

Solve for the value of $y$. Note that $y_1=y-3$
$$begin{aligned}
tan84degree&=dfrac{y_1}{d_1}\
tan84degree&=dfrac{y-3}{18}\
end{aligned}$$

Using Transposition Method.
$$begin{aligned}
tan84degree(18)&=y-3\
171.26&=y-3\
171.26+3&=y\
174.26&=y
end{aligned}$$

Solving for $y_2$.
$$begin{aligned}
y_2&=y-6\
y_2&=174.26-6\
y_2&=168.26
end{aligned}$$

Solving for the value of $theta_2$ using the formula in **Step 5**.
$$begin{aligned}
tantheta_2&=dfrac{y_2}{185}\
end{aligned}$$

Substituting the value of $y_2$ into the equation.
$$begin{aligned}
tantheta_2&=dfrac{168.26}{185}\
tantheta_2&=0.909\
theta_2&=arctan(0.909)\
theta_2&=42.29degree
end{aligned}$$

b. The tangent ratio is the opposite side divided by the adjacent rectangular side:

$$
tan{11text{textdegree}}=dfrac{120}{x}
$$

Multiply both sides of the equation by $x$:

$$
xtan{11text{textdegree}}=120
$$

Divide both sides of the equation by $tan{11text{textdegree}}$:

$$
x=dfrac{120}{tan{11text{textdegree}}}approx 617
$$

The minimum distance is then $x$ increase by the clearance needed:

$$
617+50=667ft
$$

a. The sum of all angles in a triangle is 180$text{textdegree}$:

$$
180text{textdegree}-90text{textdegree}-67text{textdegree}=23text{textdegree}
$$

The cosine ratio is the adjacent rectangular side divided by the hypotenuse.

$$
cos{23text{textdegree}}=dfrac{18}{x}
$$

Multiply both sides of the equation by $x$:

$$
xcos{23text{textdegree}}=18
$$

Divide both sides of the equation by $cos{23text{textdegree}}$:

$$
x=dfrac{18}{cos{23text{textdegree}}}approx dfrac{18}{0.921}approx 19.55
$$

b. The sine ratio of an angle is equal to the cosine ratio of the complementary angle:

$$
sin{67text{textdegree}}=cos{23text{textdegree}}approx 0.921
$$

a. $cos{23text{textdegree}}=dfrac{18}{x}$

b. $sin{67text{textdegree}}=cos{23text{textdegree}}approx 0.921$

a. Factor out the greatest common factor:

$$
2x^2-2x-4=2(x^2-x-2)
$$

Factor further:

$$
2x^2-2x-4=2(x^2-x-2)=2(x+1)(x-2)
$$

b. Factor out the greatest common factor:

$$
4(x^2-6x+9)
$$

Use $a^2-2ab+b^2=(a-b)^2$:

$$
4(x-3)^2
$$

a. $2(x+1)(x-2)$

b. $4(x-3)^2$

a. The $y$-intercept is the point at which the $x$-coordinate is zero and is thus equal to the constant term of the polynomial: $y=-8$.

b. The $x$-intercepts are the intersections of the graph with the $x$-axis and thus are at $x=-2$ and $x=4$.

c. The lowest point on the graph is at $(1,-9)$.

a. $y=-8$

b. $x=-2$ and $x=4$

c. $(1,-9)$

a. If the probability of selecting a boy is $dfrac{1}{3}$, then one third of the students in the class should be boys:

$$
36cdot dfrac{1}{3}=12
$$

Thus there are 12 boys in the class.

b. f the probability of selecting a boy is $dfrac{1}{3}$, then two thirds of the students in the class should be girls and thus there are twice as much girls as boys:

$$
2cdot 12=24
$$

Thus there are 24 girls in the class.

c. Since the total probability is 100%=1, the probability of selecting a girl is 1 decreased by the probability of selecting a boy:

$$
P(girl)=1-P(boy)=1-dfrac{1}{3}=dfrac{2}{3}
$$

d. If the probability of selecting a boy is $dfrac{1}{3}$, then one third of the students in the class should be boys:

$$
24cdot dfrac{1}{3}=8
$$

Thus the class has 8 boys and 16 girls. After one boy has been selected, we can then choose 7 boys out of 23 students:

$$
P(2^{nd} boy)=dfrac{7}{23}
$$

a. 12 boys

b. 24 girls

c. $dfrac{2}{3}$

d. $dfrac{7}{23}$

The upper cell contains the product of the left and right cell, the lower cell contains the sum of the left and right cell.

a. We need to find two numbers whose product is $-80$ and whose sum is 2. Those numbers are then 10 and $-8$. These numbers should be filled in in the left and right cell but the order does not matter.

b. We need to find two numbers whose product is $12$ and whose sum is $-7$. Those numbers are then $-4$ and $-3$. These numbers should be filled in in the left and right cell but the order does not matter.

c. We need to find two numbers whose product is $0$ and whose sum is 7. Those numbers are then 0 and $7$. These numbers should be filled in in the left and right cell but the order does not matter.

d. We need to find two numbers whose product is $-81$ and whose sum is 0. Those numbers are then $-9$ and $9$. These numbers should be filled in in the left and right cell but the order does not matter.

e. We need to find two numbers whose product is $6x^2$ and whose sum is $5x$. Those numbers are then $3x$ and $2x$. These numbers should be filled in in the left and right cell but the order does not matter.

f. We need to find two numbers whose product is $-7x^2$ and whose sum is $-6x$. Those numbers are then $-7x$ and $x$. These numbers should be filled in in the left and right cell but the order does not matter.

a. 10 and $-8$

b. $-4$ and $-3$

c. 0 and 7

d. $-9$ and 9

e. $3x$ and $2x$

f. $-7x$ and $x$