Difference of Squares:$a^2-b^2=(a-b)(a+b)$

Perfect trinomial squares: $a^2+2ab+b^2=(a+b)^2$

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

b. $3x^2+15x+12=3(x^2+5x+4)=3(x+4)(x+1)$

c. Perfect trinomial squares: $2x^2+4x+2=2(x^2+2x+1)=2(x+1)^2$

d. Difference of squares: $x^2-36=(x-6)(x+6)$

e. Perfect trinomial squares: $x^2+2ax+a^2=(x+a)^2$

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

g. see (a) to (f)

a. $(x+2)(x+4)$

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

c. $2(x+1)^2$

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

e. $(x+a)^2$

f. $(x-y)(x+y)$

g. see (a) to (f)

The graph is a parabola with axis of symmetry at $x=0$ and vertex at $(-4,0)$. The function has an $x$-intercept at $x=-2$ and $x=2$, while it has a $y$-intercept at $y=-4$. The function has a minimum at $(0,-4)$. The function is negative at $-2<x<2$ and positive at $x2$. The function is increasing at $x>0$ and decreasing at $x<0$.

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

$$
sin{15text{textdegree}}=dfrac{x}{20}
$$

Multiply both sides of the equation by 20:

$$
5.18approx 20sin{15text{textdegree}}=x
$$

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

$$
tan{15text{textdegree}}=dfrac{5}{x}
$$

Multiply both sides of the equation by $x$:

$$
xtan{15text{textdegree}}=5
$$

Divide both sides of the equation by $sin{15text{textdegree}}$:

$$
x=dfrac{5}{tan{15text{textdegree}}}approx 18.66
$$

c. Use the Pythagorean theorem:

$$
x^2=6^2+12^2=180
$$

Take the square root of both sides of the equation:

$$
x=6sqrt{5}approx 13.42
$$

a. $xapprox 5.18$

b. $xapprox 18.66$

c. $xapprox 13.42$

Corresponding sides in similar triangles have the same proportions:

$$
dfrac{x}{6}=dfrac{12}{8}=dfrac{3}{2}
$$

Multiply both sides of the equation by $6$:

$$
x=dfrac{3cdot 6}{2}=9
$$

Determine $y$:

$$
dfrac{y}{6}=dfrac{8}{12}=dfrac{2}{3}
$$

Multiply both sides of the equation by $6$:

$$
y=dfrac{2cdot 6}{3}=4
$$

Determine $z$:

$$
dfrac{z}{10}=dfrac{8}{12}=dfrac{2}{3}
$$

Multiply both sides of the equation by $10$:

$$
z=dfrac{2cdot 10}{3}=dfrac{20}{3}approx 6.67
$$

$x=9$, $y=4$ and $z=dfrac{20}{3}$

a. The probability is the number of favorable outcomes divided by the total number of outcomes:

$$
P(20+older)=dfrac{250+250+250}{1000}=dfrac{750}{1000}=0.75=75%
$$

b. Use the addition rule:

$$
P(under 20text{ or } A)=P(under 20)+P(A)-P(under 20text{ and }A)
$$

$$
=dfrac{250}{1000}+dfrac{371}{1000}-dfrac{152}{1000}=dfrac{469}{1000}=46.9%
$$

Since the expression contains OR, we know that the event is a union.

c. The complement of an event is 100% decreased by the probability of the event:

$$
P(20+oldertext{ and }notA)=P(nottext{ under} 20text{ and }not A)=1-P(under 20text{ or } A)
$$

$$
=1-dfrac{469}{1000}=dfrac{531}{1000}=53.1%
$$

We prove that $overline KL cong overline QP$ using the flowchart proof as follows:


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

$$
P(rap)=dfrac{4+1}{3+6+4+5+1+3}=dfrac{5}{22}approx 0.227=22.7%
$$

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

$$
P(traditional)=dfrac{3+6+3}{3+6+4+5+1+3}=dfrac{12}{22}=dfrac{6}{11}approx 0.545=54.5%
$$

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

$$
P(traditional: pop)=dfrac{3}{3+6+4+5+1+3}=dfrac{3}{22}approx 0.136=13.6%
$$

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

$$
P(not: country)=dfrac{6+4+1+3}{3+6+4+5+1+3}=dfrac{14}{22}=dfrac{7}{11}approx 0.636=63.6%
$$

a. Replace the letters with the corresponding letters given in the congruence of the triangles:

$$
overline{DV}cong overline{RS}
$$

$$
mangle RYS=mangle DNV
$$

b. A bisector divides an angle into two equal angles:

$$
angle DABcong angle BAC
$$

c. If two angles have the same measure in a triangle, then the opposite sides of the angle are congruent (have the same length):

$$
overline{WY}cong overline{QY}
$$

d. Adjacent angles in a parallegram are supplementary:

$$
mangle C=180text{textdegree}-148 text{textdegree}=32text{textdegree}
$$

a. $overline{DV}cong overline{RS}$, $mangle RYS=mangle DNV$

b. $angle DABcong angle BAC$

c. $overline{WY}cong overline{QY}$

d. $32text{textdegree}$