$1cdot a=a$

$$
adiv 1=a
$$

#### (a)

In this case, Mr. Wonder is correct because we can simplify the given expression and get $1$ as a result, so, it is correct that

$$
dfrac{16x}{16x}=1, xne0
$$

It must be $xne0$ because the fraction is defined when the denominator is not equal to $0$, in this case, we have:

$$
16xne0Leftrightarrow xne 0
$$

#### (b)

$x$ can not be equal to zero because division by zero is not defined.

#### (c)

We can conclude that this expression is equal to $1$, and it is defined when

$$
x-3ne 0Rightarrow xne3
$$

So, we have to exclude $3$.

#### (d)

For example, consider the following expression:

$$
dfrac{x+5}{x+5}=1,
$$

and it is defined when

$$
x+5ne 0Rightarrow xne -5
$$

#### (e)

In this case, Mr. Wonder is correct. We can get $dfrac{x}{y}$ again after simplifying if we multiply it by a fraction which is equal to $1$.

a) He is correct; b) Division by $0$ is not defined; c) $dfrac{x-3}{x-3}=1, xne 3; d) dfrac{x+5}{x+5}=1, x ne -5;$ e) He is correct

textbf{(a)} Simplifying $dfrac{x^2}{x^2}$:
begin{align*}
dfrac{textcolor{red}{x^2}}{textcolor{red}{x^2}}&=1
intertext{textbf{(b)} Simplifying $dfrac{x}{x} cdot dfrac{x}{x} cdot dfrac{x}{3} $:}
dfrac{textcolor{red}{x}}{textcolor{red}{x}} cdot dfrac{textcolor{red}{x}}{textcolor{red}{x}} cdot dfrac{x}{3}&= 1 cdot 1 cdot dfrac{x}{3}\
&=dfrac{x}{3}
intertext{Denominator is never zero}
intertext{textbf{(c)} Simplifying $dfrac{x-2}{x-2} cdot dfrac{x+5}{x-1} $:}
dfrac{textcolor{red}{x-2}}{textcolor{red}{x-2}} cdot dfrac{x+5}{x-1}&=dfrac{x+5}{x-1}
intertext{Denominator is zero for $x-1=0 rightarrow x=1$}
intertext{textbf{(d)} Simplifying $dfrac{9}{x} cdot dfrac{x}{9} $:}
dfrac{textcolor{red}{9}}{textcolor{blue}{x}} cdot dfrac{textcolor{blue}{x}}{textcolor{red}{9}}&=1
intertext{textbf{(e)} Simplifying $dfrac{h cdot h cdot k}{h} $:}
dfrac{textcolor{red}{h} cdot h cdot k}{textcolor{red}{h}} &=h cdot k
intertext{textbf{(f)} Simplifying $dfrac{(2m-5)(m+6)}{(m+6)(3m+1)}$:}
dfrac{(2m-5)textcolor{red}{(m+6)}}{textcolor{red}{(m+6)}(3m+1)}&=dfrac{2m-5}{3m+1}\
intertext{Denominator is zero for $3m+1=0 rightarrow m=-dfrac{1}{3}$}
end{align*}

begin{align*}
intertext{textbf{(g)} Simplifying $dfrac{6(n-2)^2}{3(n-2)} $:}
dfrac{6textcolor{red}{(n-2)}^2}{3textcolor{red}{(n-2)}}&=2(n-2)\
&=2n-4
intertext{textbf{(h)} Simplifying $dfrac{3-2x}{(4x-2)(3-2x)} $:}
dfrac{textcolor{red}{3-2x}}{(4x-1)textcolor{red}{(3-2x)}}&=dfrac{1}{4x-1}
intertext{Denominator is zero for $4x-1=0 rightarrow x=dfrac{1}{4}$}
end{align*}

(a) 1 (b) $dfrac{x}{3}$ (c) $dfrac{x+5}{x-1}$ (d) 1 (e) $h cdot k$ (f) $2n-4$ (h) $dfrac{1}{4x-1}$

$dfrac{4x}{x}$

$$
dfrac{4+x}{x}
$$

$$
dfrac{x}{x}=1Rightarrow dfrac{4x}{x}=4
$$

$x=2$

$dfrac{4x}{x}=dfrac{4(2)}{2}=dfrac{8}{2}=4checkmark$

$x=5$

$dfrac{4x}{x}=dfrac{4(5)}{5}=dfrac{20}{5}=4checkmark$

$x=0$

$dfrac{4x}{x}=dfrac{4(0)}{0}=dfrac{0}{0}$ (undefined!!!)

He is not correct because for $x=0$ the expression is undefined.

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

$x=1$

$dfrac{4+x}{x}=dfrac{4+1}{1}=5checkmark$

$x=0$

$dfrac{4+x}{x}=dfrac{4+0}{0}=dfrac{4}{0}$ (undefined!!!)

$x=2$

$dfrac{4+x}{x}=dfrac{4+2}{2}=3not=5$

$$
dfrac{4+x}{x}=dfrac{4}{x}+dfrac{x}{x}=dfrac{4}{x}+1
$$

$$
dfrac{x^2+x+3}{x+3}rightarrow x^2
$$

The fraction cannot be simplified by $x+3$ because the denominator cannot be factored by $x+3$. Therefore the statement is false.

$$
dfrac{(x+2)(x+3)}{x+3}rightarrow x+2
$$

$$
dfrac{(x+2)(cancel{x+3})}{cancel{x+3}}=x+2checkmark
$$

We can simplify the fraction by $x+3$ for $xnot=-3$:

#### (a)

In this case, the denominator is not equal to $0$ when:

$$
x^2-9ne 0Rightarrow (x-3)(x+3)ne 0Rightarrow x=pm3
$$

We will now simplify it on the following way:

$$
dfrac{x^2+6x+9}{(x-3)(x+3)}=dfrac{(x+3)^2}{(x-3)(x+3)}=dfrac{x+3}{x+3}cdotdfrac{x+3}{x-3}=1cdot dfrac{x+3}{x-3}=dfrac{x+3}{x-3}
$$

Now, we will substitute $5$ for $x$ in order to verify simplified expression:

$$
dfrac{5^2+6cdot 5+9}{5^2-9}=dfrac{64}{16}=4, dfrac{5+3}{5-3}=dfrac{8}{2}=4
$$

#### (b)

In this case, the denominator is not equal to $0$ when:

$$
3x^2+7x+2ne 0 Rightarrow 3(x+2)left(x+dfrac{1}{3} right)=0Rightarrow xne -2, xne -dfrac{1}{3}
$$

We will now simplify it on the following way:

$$
dfrac{2x^2-x-10}{3x^2+7x+2}=dfrac{2(x+2)left(x-tfrac{5}{2} right)}{3(x+2)left(x+tfrac{1}{3} right)}=dfrac{2left(x-tfrac{5}{2} right)}{3left( x+tfrac{1}{3}right)}cdot1cdotdfrac{x+2}{x+2}=dfrac{2left(x-tfrac{5}{2} right)}{3left( x+tfrac{1}{3}right)}cdot1=dfrac{2left(x-tfrac{5}{2} right)}{3left( x+tfrac{1}{3}right)}=dfrac{2x-5}{3x+1}
$$

Now, we will substitute $1$ for $x$ in order to verify simplified expression:

$$
dfrac{2cdot 1^2-1-10}{3cdot 1^2+7+2}=-dfrac{9}{12}=-dfrac{3}{4}, dfrac{2cdot1-5}{3cdot1+1}=-dfrac{3}{4}
$$

#### (c)

In this case, the denominator is not equal to $0$ when:

$$
28x^2-x-15=0Rightarrow 28(x-tfrac{3}{2})(x+tfrac{5}{7})
$$

We will now simplify it on the following way:

$$
dfrac{28x^2-x-15}{28x^2-x-15}=1
$$

Now, we will substitute $1$ for $x$ in order to verify simplified expression:

$$
dfrac{28cdot 1-1-15}{28cdot 1-1-15}=dfrac{12}{12}=1
$$

#### (d)

In this case, the denominator is not equal to $0$ when:

$$
2x+8ne 0Rightarrow xne -4
$$

We will now simplify it on the following way:

$$
dfrac{x^2+4x}{2x+8}=dfrac{x(x+4)}{2(x+4)}=dfrac{x}{2}cdot dfrac{x+4}{x+4}=dfrac{x}{2}cdot 1=dfrac{x}{2}
$$

Now, we will substitute $1$ for $x$ in order to verify simplified expression:

$$
dfrac{1+4}{2+8}=dfrac{5}{10}=dfrac{1}{2}, dfrac{1}{2}
$$

a) $dfrac{x+3}{x-3}$; b) $dfrac{2x-5}{3x+1}$; c) $1$; d) $dfrac{x}{2}$

In order to rewrite rational expressions in simpler forms, we need to rewrite the numerator and the denominator applying, mostly, square binomial or difference of squares, and in that way we get multiplying expressions in the denominator and numerator.

In that case, we can cancel the same expressions and in that way simplify the given rational expression.

For example:

$$
dfrac{x^2-4}{x+2}=dfrac{(x-2)(x+2)}{x+2}=dfrac{x-2}{1}cdotdfrac{x+2}{x+2}=(x-2)cdot1=x-2
$$

a-

$dfrac {x^2-8x+16}{3x^2-10x-8}$          For $x neq -dfrac {2}{3}$ or 4          (Given)

$dfrac {(x-4)(x-4)}{(3x+2)(x-4)}$          For $x neq -dfrac {2}{3}$ or 4          (Factorizing)

$dfrac {(x-4) cancel{(x-4)}}{(3x+2) cancel {(x-4)}}$

$dfrac {x-4}{3x+2}$         For $x neq -dfrac {2}{3}$ or 4

b-

$dfrac {10x+25}{2x^2-x-15}$          For $x neq -dfrac {5}{2}$ or 3          (Given)

$dfrac {5(2x+5)}{(x-3)(2x+5)}$          For $x neq -dfrac {5}{2}$ or 3          (Factorizing)

$dfrac {5 cdot cancel{(2x+5)}}{(x-3) cancel {(2x+5)}}$

$dfrac {5}{x-3}$         For $x neq -dfrac {5}{2}$ or 3

c-

$dfrac {(k-4)(2k+1)}{5(2k+1)} div dfrac {(k-3)(k-4)}{10(k-3)}$          For $k neq 3, 4, textrm { ~or ~} -dfrac {1}{2}$          (Given)

$dfrac {(k-4) cancel {(2k+1)}}{5 cdot cancel {(2k+1)}} times dfrac {10 cdot cancel {(k-3)}}{ cancel {(k-3)}(k-4)}$          For $k neq 3, 4, textrm { ~or ~} -dfrac {1}{2}$

$dfrac {(k-4) }{5} times dfrac {10}{(k-4)}$          For $k neq 3, 4, textrm { ~or ~} -dfrac {1}{2}$

$dfrac { cancel {(k-4)} }{5} times dfrac {10}{ cancel {(k-4)}}$          For $k neq 3, 4, textrm { ~or ~} -dfrac {1}{2}$

$=dfrac {10}{5}=2$          For $k neq 3, 4, textrm { ~or ~} -dfrac {1}{2}$

a-

$dfrac {x^2-8x+16}{3x^2-10x-8}$          For $x neq -dfrac {2}{3}$ or 4          (Given)

$=dfrac {x-4}{3x+2}$         For $x neq -dfrac {2}{3}$ or 4

b-

$dfrac {10x+25}{2x^2-x-15}$          For $x neq -dfrac {5}{2}$ or 3          (Given)

$=dfrac {5}{x-3}$         For $x neq -dfrac {5}{2}$ or 3

c-

$dfrac {(k-4)(2k+1)}{5(2k+1)} div dfrac {(k-3)(k-4)}{10(k-3)}$          For $k neq 3, 4, textrm { ~or ~} -dfrac {1}{2}$          (Given)

$=dfrac {10}{5}=2$          For $k neq 3, 4, textrm { ~or ~} -dfrac {1}{2}$

$$
=dfrac{9}{3}timesdfrac{2}{14}=dfrac{3}{7}
$$

(a) Reorder the terms on the numerator to convert one fraction term to an integer and simplify remaining fractional term by dividing both sides by common factor 2. Then multiply integer by numerator of remaining fraction $;$

$$
=dfrac{3}{5}timesdfrac{25}{12}=dfrac{5}{4}
$$

(b) $;$

#### (a)

In this case, the initial value is $P=$1000, r=0.03, t=1$, so, continuously compounded interest will be:

$$
A=Pe^{rt}=10 000cdot e^{0.03cdot1}=10 304.545
$$

We can notice that amount of money we got is very close to the hourly compounding.

#### (b)

First, we will consider continuously compounded interest and get:

$$
A=10 000 000 000e^{0.08cdot1}=10832870 676.75
$$

Now, we will consider compound interest which is compounded daily, so, we get:

$$
P=10000000000left(1+dfrac{0.08}{365} right)^{365cdot1}=10832775717.93
$$

We can notice that the difference between total amounts in those cases is around $$94 958.82$.

#### (c)

The conclusion is that big investors might benefit from continuous compounding rather than daily compounding.

a) $$10 304.545$; b) $$10832870 676.75, $10832775717.93$; c) Big investors.

#### (a)

We can expand the given binominal in the following way:

$$
(a+b)^3=(a+b)^2(a+b)=(a^2+2ab+b^2)(a+b)=a^3+a^2b+2a^2b+2ab^2+ab^2+b^3=a^3+3a^2b+3ab^2+b^3
$$

#### (b)

Using part a), we can expand the given expression in the following way:

$$
(2m+5)^3=(2m)^3+3cdot(2m)^2cdot 5+3cdot 2mcdot 5^2+5^3=8m^3+60m^2+150m+125
$$

a) $a^3+3a^2b+3ab^2+b^3$; b) $8m^3+60m^2+150m+125$

$$
f(x)=(x-1)^2(x-3)^3(x-5)^2
$$

$(x-1)^2=0Rightarrow x_1=1$ (multiplicity 2) (the graph touches the $x$-axis)

$(x-3)^3=0Rightarrow x_2=3$ (multiplicity 3) (the graph crosses the $x$-axis)

$(x-5)^2=0Rightarrow x_3=5$ (multiplicity 2) (the graph touches the $x$-axis)

$$
g(x)=-(x-1)^2(x-3)^3(x-5)^2
$$

$(x-1)^2=0Rightarrow x_1=1$ (multiplicity 2) (the graph touches the $x$-axis)

$(x-3)^3=0Rightarrow x_2=3$ (multiplicity 3) (the graph crosses the $x$-axis)

$(x-5)^2=0Rightarrow x_3=5$ (multiplicity 2) (the graph touches the $x$-axis)

$$
g(x)=-f(x)
$$

We express $g(x)$ in terms of $f(x)$:

$BC^2=AB^2+AC^2-2cdot ABcdot ACcdot cos A$

$75^2=300^2+250^2-2cdot 300cdot 250cdot cos A$

$75^2=300^2+250^2-2cdot 300cdot 250cdot cos A$

$5625=90,000+62,500-150,000cos A$

$150,000cos A=146,875$

$cos A=dfrac{146,875}{150,000}$

$cos Aapprox 0.9792$

$$
Aapprox 11.71text{textdegree}
$$

We apply the Law of Cosines to determine the angle $A$:

The plane should change its course by $11.71text{textdegree}$.

In order to calculate $sin theta$, we will use the following identity:

$$
sin^2theta+cos^2theta=1
$$

$$
sin^2theta+left(dfrac{8}{17} right)^2=1
$$

$$
sin^2theta+dfrac{64}{289}=1
$$

$$
sin^2theta=1-dfrac{64}{289}=dfrac{289-64}{289}
$$

$$
sin^2theta=dfrac{225}{289}Rightarrow sin theta=pmsqrt{dfrac{225}{289}}=pmdfrac{15}{17}
$$

$sintheta=pmdfrac{15}{17}$

Given $dfrac{3x^{2}+11x-4}{2x^{2}+11x+12}$ we factor the terms by grouping; the top becomes $3x^{2}+12x-x-4=3x(x+4)-(x+4)=(3x-1)(x+4)$. The denominator factors as; $2x^{2}+3x+8x+12=2x(x+4)+3(x+4)=(2x+3)(x+4)$ , then the rational expression becomes $dfrac{(3x-1)(x+4)}{(2x+3)(x+4)}$. So we have a ‘Giant 1’ where $dfrac{x+4}{x+4}=1$ and we simplify rational to $dfrac{(3x-1)}{(2x+3)}$. The denominator is zero; $2x+3=0$ when $x=-dfrac{3}{2}$ and since $x+4$ appears on both sides $x+4=0$ when $x=-4$. Therefore $x=-dfrac{3}{2}$ and $x=-4$ are excluded values from the domain. Only $x=-dfrac{3}{2}$ is an asymptote.

$x=-4$ and $x=-dfrac{3}{2}$ are excluded

$dfrac {2}{3} cdot {9}{14}$

$=dfrac {cancel {2}}{cancel {3}} cdot dfrac {cancel {3} cdot 3}{cancel {2} cdot 7}$

$=dfrac {3}{7}$

$dfrac {3}{5} div dfrac {12}{25}$

$=dfrac {3}{5} times dfrac {25}{12}$

$=dfrac {cancel {3}}{cancel {5}} times dfrac {cancel {5} cdot 5}{cancel {3} cdot 4}$

$=dfrac {5}{4}$

$dfrac {2}{3} cdot {9}{14}=dfrac {3}{7}$

$dfrac {3}{5} div dfrac {12}{25}=dfrac {5}{4}$

$$
dfrac{4x+3}{x-5}cdot dfrac{x-5}{x+3}
$$

$$
left{-3, 5right}
$$

The values of $x$ which must be excluded are:

$$
dfrac{4x+3}{cancel{x-5}}cdot dfrac{cancel{x-5}}{x+3}=dfrac{4x+3}{x+3}
$$

$$
dfrac{x+2}{9x-1}+dfrac{2x+1}{9x-1}
$$

$$
left{dfrac{1}{9}right}
$$

The values of $x$ which must be excluded are:

$$
dfrac{x+2}{9x-1}+dfrac{2x+1}{9x-1}=dfrac{x+2+2x+1}{9x-1}=dfrac{3x+3}{9x-1}
$$

$$
dfrac{2m+3}{3m-2}cdotdfrac{7+4m}{3+2m}
$$

$$
left{-dfrac{3}{2}, dfrac{2}{3}right}
$$

The values of $m$ which must be excluded are:

$dfrac{2m+3}{3m-2}cdotdfrac{7+4m}{3+2m}$

$$
=dfrac{cancel{2m+3}}{3m-2}cdotdfrac{4m+7}{cancel{2m+3}}=dfrac{4m+7}{3m-2}
$$

$$
dfrac{(y-2)^3}{3y}cdot dfrac{y+5}{(y+2)(y-2)}
$$

$$
{-2, 0, 2}
$$

The values of $y$ which must be excluded are:

$dfrac{(y-2)^3}{3y}cdot dfrac{y+5}{(y+2)(y-2)}$

$=dfrac{(y-2)^2(cancel{y-2})}{3y}cdot dfrac{y+5}{(y+2)(cancel{y-2})}$

$$
=dfrac{(y-2)^2(y+5)}{3y(y+2)}
$$

$$
dfrac{15x^3}{3y}+dfrac{10x^2y}{4y^2}
$$

$$
ynot=0
$$

The values of $x,y$ which must be excluded are:

$dfrac{15x^3}{3y}+dfrac{10x^2y}{4y^2}$

$dfrac{5x^3}{y}+dfrac{5x^2}{2y}$

$=dfrac{5x^3cdot 2}{2y}+dfrac{5x^2}{2y}$

$$
=dfrac{10x^3+5x^2}{2y}
$$

$$
dfrac{(5x-2)(3x+1)}{(2x-3)^2}+dfrac{(5x-2)(x-4)}{(x-4)(2x-3)}
$$

$$
left{dfrac{3}{2}, 4right}
$$

The values of $x$ which must be excluded are:

$dfrac{(5x-2)(3x+1)}{(2x-3)^2}+dfrac{(5x-2)(cancel{x-4})}{(cancel{x-4})(2x-3)}$

$=dfrac{(5x-2)(3x+1)}{(2x-3)^2}+dfrac{5x-2}{2x-3}$

$=dfrac{(5x-2)(3x+1)}{(2x-3)^2}+dfrac{(5x-2)(2x-3)}{(2x-3)^2}$

$=dfrac{(5x-2)(3x+1)+(5x-2)(2x-3)}{(2x-3)^2}$

$=dfrac{(5x-2)(3x+1+2x-3)}{(2x-3)^2}$

$=dfrac{(5x-2)(5x-2)}{(2x-3)^2}$

$$
=left(dfrac{5x-2}{2x-3}right)^2
$$

$$
dfrac{20}{22}cdot dfrac{14}{35}
$$

$dfrac{20}{22}cdot dfrac{14}{35}=dfrac{2cdot 2cdot 5}{2cdot 11}cdotdfrac{2cdot 7}{5cdot 7}$

$=dfrac{cancel{2}cdot 2cdot cancel{5}}{cancel{2}cdot 11}cdotdfrac{2cdot cancel{7}}{cancel{5}cdot cancel{7}}$

$$
=dfrac{4}{11}
$$

$$
dfrac{12}{40}divdfrac{15}{6}
$$

$dfrac{12}{40}divdfrac{15}{6}$

$=dfrac{2cdot 2cdot 3}{2cdot 2cdot 2cdot 5}cdot dfrac{2cdot 3}{3cdot 5}$

$=dfrac{cancel{2}cdot cancel{2}cdot cancel{3}}{cancel{2}cdot cancel{2}cdot cancel{2}cdot 5}cdot dfrac{cancel{2}cdot 3}{cancel{3}cdot 5}$

$$
=dfrac{3}{25}
$$

$$
dfrac{5x-15}{3x^2+10x-8}divdfrac{x^2+x-12}{3x^2-8x+4}
$$

$dfrac{5x-15}{3x^2+12x-2x-8}divdfrac{x^2+4x-3x-12}{3x^2-6x-2x+4}$

$=dfrac{5(x-3)}{3x(x+4)-2(x+4)}divdfrac{x(x+4)-3(x+4)}{3x(x-2)-2(x-2)}$

$$
=dfrac{5(x-3)}{(x+4)(3x-2)}cdotdfrac{(x-2)(3x-2)}{(x+4)(x-3)}
$$

$=dfrac{5(cancel{x-3})}{(x+4)(cancel{3x-2})}cdotdfrac{(x-2)(cancel{3x-2})}{(x+4)(cancel{x-3})}$

$$
=dfrac{5(x-2)}{(x+4)^2}
$$

$$
dfrac{12x-8}{x^2-2x-15}cdotdfrac{x^2-x-12}{3x^2-9x-12}
$$

$dfrac{12x-8}{x^2-2x-15}cdotdfrac{x^2-x-12}{3x^2-9x-12}$

$=dfrac{4(3x-2)}{(x^2-2x+1)-16}cdotdfrac{x^2+3x-4x-12}{3(x^2-3x-4)}$

$=dfrac{4(3x-2)}{(x-1)^2-16}cdotdfrac{x(x+3)-4(x+3)}{3(x^2+x-4x-4)}$

$=dfrac{4(3x-2)}{(x-1-4)(x-1+4)}cdotdfrac{(x+3)(x-4)}{3[x(x+1)-4(x+1)]}$

$=dfrac{4(3x-2)}{(x-5)(x+3)}cdotdfrac{(x+3)(x-4)}{3(x-4)(x+1)}$

$dfrac{4(3x-2)}{(x-5)(cancel{x+3})}cdotdfrac{(cancel{x+3})(cancel{x-4})}{3(cancel{x-4})(x+1)}$

$$
=dfrac{4(3x-2)}{3(x-5)(x+1)}
$$

$$
dfrac{5x^2+34x-7}{10x}cdot dfrac{5x}{x^2+4x-21}
$$

$dfrac{5x^2+34x-7}{10x}cdot dfrac{5x}{x^2+4x-21}$

$=dfrac{5x^2+35x-x-7}{10x}cdotdfrac{5x}{(x^2+4x+4)-25}$

$=dfrac{5x(x+7)-(x+7)}{10x}cdot dfrac{5x}{(x+2)^2-25}$

$=dfrac{(x+7)(5x-1)}{10x}cdotdfrac{5x}{(x+2-5)(x+2+5)}$

$=dfrac{(x+7)(5x-1)}{10x}cdotdfrac{5x}{(x-3)(x+7)}$

$dfrac{(cancel{x+7})(5x-1)}{2cdot cancel{5x}}cdotdfrac{cancel{5x}}{(x-3)(cancel{x+7})}$

$$
=dfrac{5x-1}{2(x-3)}
$$

$$
dfrac{2x^2+x-10}{x^2+2x-8}divdfrac{4x^2+20x+25}{x+4}
$$

$dfrac{2x^2-4x+5x-10}{(x^2+2x+1)-9}divdfrac{(2x+5)^2}{x+4}$

$dfrac{2x(x-2)+5(x-2)}{(x+1)^2-9}divdfrac{(2x+5)^2}{x+4}$

$=dfrac{(x-2)(2x+5)}{(x+1-3)(x+1+3)}divdfrac{(2x+5)^2}{x+4}$

$=dfrac{(x-2)(2x+5)}{(x-2)(x+4)}divdfrac{(2x+5)^2}{x+4}$

$$
=dfrac{(x-2)(2x+5)}{(x-2)(x+4)}cdot dfrac{x+4}{(2x+5)^2}
$$

$dfrac{(cancel{x-2})(cancel{2x+5})}{(cancel{x-2})(cancel{x+4})}cdot dfrac{cancel{x+4}}{(cancel{2x+5})(2x+5)}$

$$
=dfrac{1}{2x+5}
$$

When multiplying two rational expressions, we multiply the first numerator by the second numerator and the product is the numerator of the result. and multiply the first denominator by the second denominator and the product is the denominator of the result. then we simplify the result.

For example.
$dfrac {3}{4} times dfrac {2}{5}$

$=dfrac {3 cdot 2}{4 cdot 5}$          (Multiply both numerators and both denominators)

$=dfrac {6}{20}=dfrac {3}{10}$          (Simplify the result)

When dividing two rational expressions we convert the divisor by exchanging the numerator and denominator and convert the division to multiplication.

For example:          $dfrac {3}{4} div dfrac {2}{5}$

$=dfrac {3}{4} times dfrac {5}{2}$          (Exchanging the numerator and denominator of the divisor and exchanging division and multiplication.)

$=dfrac {15}{8}$          (Executing the multiplication process)

a-

$dfrac {x-7}{9(2x-1)} div dfrac {(x+5)(x-7)}{6x(x+5)}$          (Given)

$=dfrac {x-7}{9(2x-1)} times dfrac {6x (x+5)}{(x+5)(x-7)}$          (Invert the divisor fracion)

$=dfrac {cancel {(x-7)}}{9(2x-1)} times dfrac {6x cdot cancel {(x+5)}}{cancel {(x+5)} cancel {(x-7)}}$          (Cancel out the common factors)

$=dfrac {6x}{9(2x-1)}$          (Simplify)

$$
=dfrac {2x}{3(2x-1)}
$$

b-

$dfrac {6x^2-x-1}{3x^2+25x+8} times dfrac {x^2+4x-32}{2x^2+7x-4}$          (Given)

$=dfrac {(3x+1)(2x-1)}{(3x+1)(x+8)} times dfrac {(x+8)(x-4)}{(2x-1)(x+4)}$          (Factorizing polynomial)

$=dfrac {cancel {(3x+1)} cancel {(2x-1)}}{cancel {(3x+1)} cancel {(x+8)}} times dfrac {cancel {(x+8)}(x-4)}{ cancel {(2x-1)}(x+4)}$          (Cancel out the common factors)

$$
=dfrac {x-4}{x+4}
$$

a-          $dfrac {x-7}{9(2x-1)} div dfrac {(x+5)(x-7)}{6x(x+5)}=dfrac {2x}{3(2x-1)}$

b-          $dfrac {6x^2-x-1}{3x^2+25x+8} times dfrac {x^2+4x-32}{2x^2+7x-4}=dfrac {x-4}{x+4}$

(a) The denominator is zero; $(x+4)(x-2)=0$ when $x=-4$ and $x=2$ so $x=-4$ and $x=2$ are excluded values from the domain. Only $x=2$ is an asymptote however since $x+4$ appears on both sides of the factored rational; $dfrac{(x+4)(x+4)}{(x+4)(x-2)}$

(b) The denominator is zero; $4(x+2)(x+2)(x-3)(x-3)(x-3)(x-3)(x-3)$ when $x=-2$ and $x=3$ so these are excluded. Multiplying out terms; $dfrac{8(x+2)(x+2)(x+2)(x-3)(x-3)(x-3)}{4(x+2)(x+2)(x-3)(x-3)(x-3)(x-3)(x-3)}=dfrac{2(x+2)}{(x-3)(x-3)}$

(a) $x=-4$ and $x=2$ (b) $x=-2$ and $x=3$

#### (a)

In order to add given fractions, we need to know that $(+)(-)=(-)$, so, now we get the following:

$$
dfrac{8}{11}+left(-dfrac{3}{11} right)=dfrac{8}{11}-dfrac{3}{11}
$$

Because both fractions has the same value of the denomiantor, we can subtract them and get the final result.

$$
dfrac{8}{11}-dfrac{3}{11}=dfrac{8-3}{11}=dfrac{5}{11}
$$

#### (b)

Because both fractions have the same value of the denominator, we can add them in the following way and get the final solution:

$$
dfrac{x}{6}+dfrac{2}{6}=dfrac{x+2}{6}
$$

#### (c)

In that case, in order to add the given fractions, we need to find $LCM$ for $3$ and $5$.

$LCM$ for $3$ and $5$ is $15$, so, we will multiply the first fraction by $5$ and the second one by $3$ and get the following:

$$
dfrac{1}{3}+dfrac{2}{5}=dfrac{1}{3}cdotdfrac{5}{5}+dfrac{2}{5}cdotdfrac{3}{3}=dfrac{5}{15}+dfrac{6}{15}=dfrac{5+6}{15}=dfrac{11}{15}
$$

a) $dfrac{5}{11}$; b) $dfrac{x+2}{6}$; c) $dfrac{11}{15}$

Let’s consider two rational numbers $dfrac{a}{b}$ and $dfrac{c}{d}$, where $bnot=0, dnot=0$.

$$
dfrac{a}{b}cdotdfrac{c}{d}=dfrac{ac}{bd}
$$

$$
dfrac{a}{b}divdfrac{c}{d}=dfrac{a}{b}cdotdfrac{d}{c}=dfrac{ad}{bc}
$$

The product is a rational number, but only if $cnot=0$, therefore the set of rational numbers is not closed under division.

$P=25,000$

$r=0.06$

$A=Pleft(1+dfrac{r}{n}right)^{nt}$

$A=25,000left(1+dfrac{0.06}{4}right)^{4t}$

$A=25,000(1.015)^{4t}$

$A=25,000[(1.015)^4]^t$

$$
textcolor{#4257b2}{A=25,000(1.06136)^t}
$$

$A=Pe^{rt}$

$A=25,000(e^{0.06})^t$

$$
textcolor{#c34632}{A=25,000(1.06184)^t}
$$

$25,000(1.06136)^t=2cdot 25,000$

$(1.06136)^t=2$

$ln (1.06136)^t=ln 2$

$tln 1.06136=ln 2$

$t=dfrac{ln 2}{ln 1.06136}$

$$
tapprox 11.64
$$

b) We solve the equation $A=2P$ in case of the interest compounded quarterly:

$25,000(1.06184)^t=2cdot 25,000$

$(1.06184)^t=2$

$ln (1.06184)^t=ln 2$

$tln 1.06184=ln 2$

$t=dfrac{ln 2}{ln 1.06184}$

$$
tapprox 11.55
$$

We solve the equation $A=2P$ in case of the interest compounded continuously:

$$
(x^4-7x^2+3x+18)div (x+2)
$$

$polyhornerscheme[x=-2]{x^4-7x^2+3x+18}$

$$
x^4-7x^2+3x+18=(x+2)(x^3-2x^2-3x+9)
$$

Quotient: $x^3-2x^2-3x+9$

Remainder: $0$

$y=2sin (x)$

$$
y=cos (x)
$$

$y=2sin (x)+cos (x)$

$$
dfrac{8}{11}+left(-dfrac{3}{11}right)
$$

$$
dfrac{8}{11}+left(-dfrac{3}{11}right)=dfrac{8+(-3)}{11}=dfrac{5}{11}
$$

$$
dfrac{x}{6}+dfrac{2}{6}
$$

$$
dfrac{x}{6}+dfrac{2}{6}=dfrac{x+2}{6}
$$

$$
dfrac{1}{3}+dfrac{2}{5}
$$

$dfrac{1}{3}+dfrac{2}{5}=dfrac{1}{3}cdotdfrac{5}{5}+dfrac{2}{5}cdotdfrac{3}{3}$

$=dfrac{5}{15}+dfrac{6}{15}=dfrac{5+6}{15}$

$$
=dfrac{11}{15}
$$

$dfrac{5}{11}$; $dfrac{x+2}{6}$; $dfrac{11}{15}$

$$
dfrac{2x}{2x^2+x-21}+dfrac{7}{2x^2+x-21}
$$

$dfrac{2x}{2x^2+x-21}+dfrac{7}{2x^2+x-21}=dfrac{2x+7}{2x^2+x-21}$

$=dfrac{2x+7}{2x^2-6x+7x-21}=dfrac{2x+7}{2x(x-3)+7(x-3)}$

$=dfrac{cancel{2x+7}}{(x-3)(cancel{2x+7})}$

$=dfrac{1}{x-3}$

$$
xnot= 3, xnot=-dfrac{7}{2}
$$

$dfrac{5x}{x^2-2x-3}-dfrac{15}{x^2-2x-3}$

$dfrac{5x}{x^2-2x-3}-dfrac{15}{x^2-2x-3}$

$=dfrac{5x-15}{x^2-2x-3}$

$=dfrac{5(x-3)}{(x^2-2x+1)-4}$

$=dfrac{5(x-3)}{(x-1)^2-4}$

$=dfrac{5(x-3)}{(x-1-2)(x-1+2)}$

$=dfrac{5(cancel{x-3})}{(cancel{x-3})(x+1)}$

$=dfrac{1}{x+1}$

$$
xnot=-1,xnot=3
$$

$dfrac{3x+9}{8x^2-50}-dfrac{x+4}{8x^2-50}$

$dfrac{3x+9}{8x^2-50}-dfrac{x+4}{8x^2-50}$

$=dfrac{3x+9}{2(4x^2-25)}-dfrac{x+4}{2(4x^2-25)}$

$=dfrac{3x+9-x-4}{2(2x-5)(2x+5)}$

$=dfrac{cancel{2x+5}}{2(2x-5)(cancel{2x+5})}$

$=dfrac{1}{2(2x-5)}$

$$
xnot=dfrac{5}{2}, xnot=-dfrac{5}{2}
$$

$$
dfrac{x^2+5x-2}{3x^2+2x-8}+dfrac{2x^2-3x-6}{3x^2+2x-8}
$$

$dfrac{x^2+5x-2}{3x^2+2x-8}+dfrac{2x^2-3x-6}{3x^2+2x-8}$

$=dfrac{x^2+5x-2+2x^2-3x-6}{3x^2+2x-8}$

$=dfrac{3x^2+2x-8}{3x^2+2x-8}$

$=dfrac{3x^2+6x-4x-8}{3x^2+6x-4x-8}$

$=dfrac{3x(x+2)-4(x+2)}{3x(x+2)-4(x+2)}$

$=dfrac{(cancel{x+2})(cancel{3x-4})}{(cancel{x+2})(cancel{3x-4})}$

$=1$

$$
xnot=-2, xnot=dfrac{4}{3}
$$

#### (a)

In order to add the given fractions, we need to find $LCM$ for expressions in the denominators.

$LCM$ for $(3x+1)$ and $(x-5)(3x+1)$ is $(x-5)(3x+1)$, so, we need to multiply first fraction by $x-5$, so, we get the following:

$$
dfrac{x}{3x+1}cdotdfrac{x-5}{x-5}+dfrac{2x^2-2}{(x-5)(3x+1)}=dfrac{x(x-5)+2x^2-2}{(3x+1)(x-5)}=dfrac{x^2-5x+2x^2-2}{(3x+1)(x-5)}=dfrac{3x^2-5x-2}{(3x+1)(x-5)}=
$$

$$
=dfrac{3(x-2)(x+tfrac{1}{3})}{(3x+1)(x-5)}=dfrac{(3x+1)(x-2)}{(3x+1)(x-5)}=dfrac{x-2}{x-5}, xne5, xne -dfrac{1}{3}
$$

#### (b)

In order to add the given fractions, we need to find $LCM$ for expressions in the denominators.

$LCM$ for $(x+3)(x-3)$ and $(x+3)$ is $(x+3)(x-3)$, so, we need to multiply second fraction by $x-3$, so, we get the following:

$$
dfrac{9-3x}{(x+3)(x-3)}+dfrac{2x}{x+3}cdotdfrac{x-3}{x-3}=dfrac{9-3x+2x(x-3)}{(x+3)(x-3)}=dfrac{9-3x+2x^2-6x}{(x+3)(x-3)}=
$$

$$
=dfrac{2x^2-9x+9}{(x+3)(x-3)}=dfrac{2(x-3)(x-tfrac{3}{2})}{(x+3)(x-3)}=dfrac{(x-3)(2x-3)}{(x+3)(x-3)}=dfrac{2x-3}{x+3}, xne pm3
$$

a) $dfrac{x-2}{x-5}$; b) $dfrac{2x-3}{x+3}$

$$
dfrac{x+2}{x-4}
$$

When $x=4$ the denominator is zero and the fraction is undefined.

$3x+1=0Rightarrow x_1=-dfrac{1}{3}$

$$
x-5=0Rightarrow x_2=5
$$

b) For the exercise $28a$, the excluded values are:

$x+3=0Rightarrow x_1=-3$

$$
x-3=0Rightarrow x_2=3
$$

For the exercise $28b$, the excluded values are:

$$
dfrac{x-1}{(x+6)(3x-1)}
$$

c) We create an expression that has the excluded values of $xnot=-6$ and $xnot=dfrac{1}{3}$ by placing the factors $(x+6)$ and $left(x-dfrac{1}{3}right)$ in the denominator:

$textcolor{#4257b2}{text{Adding and Subtracting Rational Expressions}}$

First we factor denominators. Then we determine the domain of the expression, eliminating from the domain the zeros of the denominators.

We determine the Least Common Denominator by multiplying all factors of the denominators at the greatest exponent.

Then we multiply each rational expression by a fraction of the form $dfrac{p(x)}{p(x)}$ where $p(x)$ is obtained by dividing the Least Common Denominator by the denominator of the rational expression.

$dfrac{x+2}{x^2-x}-dfrac{x}{x^2-1}$

$=dfrac{x+2}{x(x-1)}-dfrac{x}{(x-1)(x+1)}$

$$
(-infty,-1)cup(-1,0)cup(0,1)cup(1,infty)
$$

$$
x(x-1)(x+1)
$$

$dfrac{x+2}{x(x-1)}-dfrac{x}{(x-1)(x+1)}$

$=dfrac{x+2}{x(x-1)}cdotdfrac{x+1}{x+1}-dfrac{x}{(x-1)(x+1)}cdotdfrac{x}{x}$

$=dfrac{(x+2)(x+1)-xcdot x}{x(x-1)(x+1)}$

$=dfrac{x^2+x+2x+2-x^2}{x(x-1)(x+1)}$

$$
=dfrac{3x+2}{x(x-1)(x+1)}
$$

In order to add the given fractions, we need to find $LCM$ for expressions in the denominators.

$LCM$ for $(x-1)$ and $(x+5)$ is $(x-1)(x+5)$, so, we need to multiply first fraction by $(x+5)$ and the second one by $(x-1)$, so, we get the following:

$$
dfrac{2x}{x-1}cdotdfrac{x+5}{x+5}+dfrac{3}{x+5}cdotdfrac{x-1}{x-1}=dfrac{2x(x+5)+3(x-1)}{(x-1)(x+5)}=dfrac{2x^2+10x+3x-3}{(x-1)(x+5)}=dfrac{2x^2+13x-3}{(x-1)(x+5)}
$$

$dfrac{2x^2+13x-3}{(x-1)(x+5)}$

#### (a)

In this case, the denominators are the same, so, we can add fractions in the following way:

$$
dfrac{5m+18}{m+3}+dfrac{4m+9}{m+3}=dfrac{5m+18+4m+9}{m+3}=dfrac{9m+27}{m+3}=dfrac{9(m+3)}{m+3}=9
$$

#### (b)

In this case, the denominators are the same, so, we can add fractions in the following way:

$$
dfrac{3a^2+a-1}{a^2-2a+1}-dfrac{2a^2-a+2}{a^2-2a+1}=dfrac{3a^2+a-1-2a^2+a-2}{a^2-2a+1}=dfrac{a^2+2a-3}{(a-1)^2}=dfrac{(a-1)(a+3)}{(a-1)^2}=dfrac{a+3}{a-1}
$$

a) $9$; b) $dfrac{a+3}{a-1}$

$$
dfrac{(x-4)^3(2x-1)}{(2x-1)(x-4)^2}
$$

$$
dfrac{cancel{(x-4)^2}(x-4)(cancel{2x-1})}{(cancel{2x-1})cancel{(x-4)^2}}=x-4
$$

$$
dfrac{7m^2-22m+3}{3m^2-7m-6}
$$

$dfrac{7m^2-22m+3}{3m^-7m-6}=dfrac{7m^2-21m-m+3}{3m^2-9m+2m-6}$

$=dfrac{7m(m-3)-(m-3)}{3m(m-3)+2(m-3)}$

$=dfrac{(cancel{m-3})(7m-1)}{(cancel{m-3})(3m+2)}$

$$
=dfrac{7m-1}{3m+2}
$$

$$
dfrac{(z+2)^9(4z-1)^7}{(z+2)^{10}(4z-1)^5}
$$

$dfrac{(z+2)^9(4z-1)^7}{(z+2)^{10}(4z-1)^5}$

$=dfrac{cancel{(z+2)^9}cancel{(4z-1)^5}(4z-1)^2}{cancel{(z+2)^9}(z+2)cancel{(4z-1)^5}}$

$$
=dfrac{(4z-1)^2}{z+2}
$$

$$
dfrac{(x+2)(x^2-6x+9)}{(x-3)(x^2-4)}
$$

$dfrac{(x+2)(x^2-6x+9)}{(x-3)(x^2-4)}$

$=dfrac{(x+2)(x-3)^2}{(x-3)(x-2)(x+2)}$

$=dfrac{(cancel{x+2})(cancel{x-3})(x-3)}{(cancel{x-3})(x-2)(cancel{x+2})}$

$$
=dfrac{x-3}{x-2}
$$

a) $x-4$;b) $dfrac{7m-1}{3m+2}$; c) $dfrac{(4z-1)^2}{z+2}$; d) $dfrac{x-3}{x-2}$

$$
dfrac{(3x-1)(x+7)}{4(2x-5)}cdotdfrac{10(2x-5)}{(4x+1)(x+7)}
$$

$dfrac{(3x-1)(cancel{x+7})}{cancel{2}cdot 2(cancel{2x-5})}cdotdfrac{cancel{2}cdot 5(cancel{2x-5})}{(4x+1)(cancel{x+7})}$

$$
=dfrac{5(3x-1)}{2(4x+1)}
$$

$$
dfrac{(m-3)(m+11)}{(2m+5)(m-3)}divdfrac{(4m-3)(m+11)}{(4m-3)(2m+5)}
$$

$dfrac{(cancel{m-3})(cancel{m+11})}{(cancel{2m+5})(cancel{m-3})}cdotdfrac{(cancel{4m-3})(cancel{2m+5})}{(cancel{4m-3})(cancel{m+11})}$

$$
=1
$$

$$
dfrac{2p^2+5p-12}{2p^2-5p+3}cdotdfrac{p^2+8p-9}{3p^2+10p-8}
$$

$dfrac{2p^2+8p-3p-12}{2p^2-2p-3p+3}cdotdfrac{(p^2+8p+16)-25}{3p^2+12p-2p-8}$

$=dfrac{2p(p+4)-3(p+4)}{2p(p-1)-3(p-1)}cdotdfrac{(p+4)^2-25}{3p(p+4)-2(p+4)}$

$=dfrac{(p+4)(2p-3)}{(p-1)(2p-3)}cdotdfrac{(p+4-5)(p+4+5)}{(p+4)(3p-2)}$

$=dfrac{(cancel{p+4})(cancel{2p-3})}{(cancel{p-1})(cancel{2p-3})}cdotdfrac{(cancel{p-1})(p+9)}{(cancel{p+4})(3p-2)}$

$$
=dfrac{p+9}{3p-2}
$$

$$
dfrac{4x-12}{x^2+3x-10}divdfrac{2x^2-13x+21}{2x^2+3x-35}
$$

$dfrac{4x-12}{x^2+3x-10}divdfrac{2x^2-13x+21}{2x^2+3x-35}$

$=dfrac{4(x-3)}{x^2+5x-2x-10}divdfrac{2x^2-6x-7x+21}{2x^2+10x-7x-35}$

$=dfrac{4(x-3)}{x(x+5)-2(x+5)}divdfrac{2x(x-3)-7(x-3)}{2x(x+5)-7(x+5)}$

$=dfrac{4(cancel{x-3})}{(cancel{x+5})(x-2)}cdotdfrac{(cancel{x+5})(cancel{2x-7})}{(cancel{x-3})(cancel{2x-7})}$

$$
=dfrac{4}{x-2}
$$

a) $dfrac{5(3x-1)}{2(4x+1)}$; b) $1$; c) $dfrac{p+9}{3p-2}$; d) $dfrac{4}{x-2}$

The x-intercepts of $g(x)$ occur at values of x for which we have $g(x)=0$. Since $g(-2)=0$ then $x=-2$ is an x-intercept. The y-intercept is $g(0)=dfrac{2}{-1}=-2$. We have division by zero if $x=1$ and since the factor $x-1$ only appears on the denominator, this is a vertical asymptote (shown below)

$$
dfrac{a}{b}+dfrac{c}{d}=dfrac{ad}{bd}+dfrac{bc}{bd}=dfrac{ab+bc}{bd}
$$

b) Let’s consider $dfrac{a}{b}, dfrac{c}{d}$ rational numbers, $bnot=0, dnot=0$. We add the two numbers:

The number $dfrac{ab+bc}{bd}$ is a rational number ($bdnot=0$ because $bnot=0,dnot=0$), therefore the set of rational numbers is closed under addition.

$f(4)=11.7$

$f(30)=3.7$

$$
f(x)=4.03sin(0.12x+1.22)+7.70
$$

$$
f(41)=4.03sin(0.12cdot 41+1.22)+7.70approx 7.12
$$

b) We determine the value of the function for $x=41$:

$$
dfrac{5}{8}+dfrac{1}{6}
$$

$$
dfrac{5}{8}cdot dfrac{3}{3}+dfrac{1}{6}cdot dfrac{4}{4}=dfrac{15}{24}+dfrac{4}{24}=dfrac{19}{24}
$$

$$
dfrac{8}{9}-dfrac{2}{3}
$$

$$
dfrac{8}{9}-dfrac{2}{3}=dfrac{8}{9}-dfrac{2}{3}cdot dfrac{3}{3}=dfrac{8}{9}-dfrac{6}{9}=dfrac{2}{9}
$$

$$
dfrac{x+5}{x+2}+dfrac{2x+1}{x+2}
$$

$dfrac{x+5}{x+2}+dfrac{2x+1}{x+2}=dfrac{x+5+2x+1}{x+2}$

$$
=dfrac{3x+6}{x+2}=dfrac{3(cancel{x+2})}{cancel{x+2}}=3
$$

$$
dfrac{x^2-3}{(x+5)(2x-1)}+dfrac{x}{2x-1}
$$

$dfrac{x^2-3}{(x+5)(2x-1)}+dfrac{x}{2x-1}$

$=dfrac{x^2-3}{(x+5)(2x-1)}+dfrac{x}{2x-1}cdotdfrac{x+5}{x+5}$

$=dfrac{x^2-3+x(x+5)}{(x+5)(2x-1)}$

$=dfrac{x^2-3+x^2+5x}{(x+5)(2x-1)}$

$=dfrac{2x^2+5x-3}{(x+5)(2x-1)}$

$=dfrac{2x^2+6x-x-3}{(x+5)(2x-1)}$

$=dfrac{2x(x+3)-(x+3)}{(x+5)(2x-1)}$

$=dfrac{(x+3)(cancel{2x-1})}{(x+5)(cancel{2x-1})}$

$$
=dfrac{x+3}{x+5}
$$

a) $dfrac{19}{24}$; b) $dfrac{2}{9}$; c) $3$; d) $dfrac{x+3}{x+5}$;

(a) The rational can be factored by grouping to give a fraction involving products; $dfrac{2x-1}{(3x+1)(x+4)}+dfrac{x+3}{(x-7)(x+4)}$

$$
=dfrac{2x-1}{(3x+1)(x+4)}timesdfrac{x-7}{x-7}+dfrac{x+3}{(x-7)(x+4)}timesdfrac{3x+1}{3x+1}
$$

$$
=dfrac{2x^{2}-15x+7}{(3x+1)(x+4)(x-7)}+dfrac{3x^{2}+10x+3}{(3x+1)(x+4)(x-7)}
$$

(b) $=dfrac{5x^{2}-5x+10}{(3x+1)(x+4)(x-7)}$

$$
=dfrac{5(x^{2}-x+2)}{(3x+1)(x+4)(x-7)}
$$

(c) The rationals can be factored to give a fraction involving products; $dfrac{2}{(x+4)}timesdfrac{x-4}{x-4}-dfrac{4x-x^{2}}{(x-4)(x+4)}$

$$
= dfrac{2x-8-4x+x^{2}}{(x+4)(x-4)}
$$

$$
= dfrac{x(x-4)+2(x-4)}{(x+4)(x-4)}
$$

$$
= dfrac{(x+2)(x-4)}{(x+4)(x-4)}
$$

$$
= dfrac{(x+2)}{(x+4)}
$$

$(a)&(b)$ $dfrac{5(x^{2}-x+2)}{(3x+1)(x+4)(x-7)}$ (c) $dfrac{(x+2)}{(x+4)}$

$$
dfrac{2x^2+x}{(2x+1)^2}-dfrac{3}{2x+1}
$$

$dfrac{2x^2+x}{(2x+1)^2}-dfrac{3}{2x+1}$

$=dfrac{x(cancel{2x+1})}{(cancel{2x+1})(2x+1)}-dfrac{3}{2x+1}$

$=dfrac{x}{2x+1}-dfrac{3}{2x+1}$

$$
=dfrac{x-3}{2x+1}
$$

$$
dfrac{x^2-3x-10}{x^2-4x-5}divdfrac{x^2-7x-18}{2x^2-5x-7}
$$

$dfrac{x^2-3x-10}{x^2-4x-5}divdfrac{x^2-7x-18}{2x^2-5x-7}$

$=dfrac{x^2-5x+2x-10}{x^2-5x+x-5}divdfrac{x^2+2x-9x-18}{2x^2+2x-7x-7}$

$=dfrac{x(x-5)+2(x-5)}{x(x-5)+(x-5)}divdfrac{x(x+2)-9(x+2)}{2x(x+1)-7(x+1)}$

$=dfrac{(cancel{x-5})(cancel{x+2})}{(cancel{x-5})(cancel{x+1})}cdot dfrac{(cancel{x+1})(2x-7)}{(cancel{x+2})(x-9)}$

$=dfrac{2x-7}{x-9}$

$$
dfrac{15x-20}{x-5}cdotdfrac{x^2-2x-15}{3x^2+5x-12}
$$

$dfrac{15x-20}{x-5}cdotdfrac{x^2-2x-15}{3x^2+5x-12}$

$=dfrac{5(3x-4)}{x-5}cdotdfrac{(x^2-2x+1)-16}{3x^2+9x-4x-12}$

$=dfrac{5(3x-4)}{x-5}cdotdfrac{(x-1)^2-16}{3x(x+3)-4(x+3)}$

$=dfrac{5(3x-4)}{x-5}cdotdfrac{(x-1-4)(x-1+4)}{(x+3)(3x-4)}$

$=dfrac{5(cancel{3x-4})}{cancel{x-5}}cdotdfrac{(cancel{x-5})(cancel{x+3})}{(cancel{x+3})(cancel{3x-4})}$

$$
=5
$$

$$
dfrac{4}{2x+3}+dfrac{x^2-x-2}{2x^2+5x+3}
$$

$dfrac{4}{2x+3}+dfrac{x^2+x-2x-2}{2x^2+2x+3x+3}$

$=dfrac{4}{2x+3}+dfrac{x^2+x-2x-2}{2x^2+2x+3x+3}$

$=dfrac{4}{2x+3}+dfrac{x(x+1)-2(x+1)}{2x(x+1)+3(x+1)}$

$=dfrac{4}{2x+3}+dfrac{(cancel{x+1})(x-2)}{(cancel{x+1})(2x+3)}$

$=dfrac{4}{2x+3}+dfrac{x-2}{2x+3}$

$=dfrac{4+x-2}{2x+3}$

$$
=dfrac{x+2}{2x+3}
$$

$dfrac{6x-4}{3x^2-17x+10}-dfrac{1}{x^2-2x-15}$

$dfrac{6x-4}{3x^2-17x+10}-dfrac{1}{x^2-2x-15}$

$=dfrac{2(3x-2)}{3x^2-15x-2x+10}-dfrac{1}{(x^2-2x+1)-16}$

$=dfrac{2(3x-2)}{3x(x-5)-2(x-5)}-dfrac{1}{(x-1)^2-16}$

$=dfrac{2(3x-2)}{(x-5)(3x-2)}-dfrac{1}{(x-1-4)(x-1+4)}$

$=dfrac{2(cancel{3x-2})}{(x-5)(cancel{3x-2})}-dfrac{1}{(x-5)(x+3)}$

$=dfrac{2}{x-5}-dfrac{1}{(x-5)(x+3)}$

$=dfrac{2}{x-5}cdotdfrac{x+3}{x+3}-dfrac{1}{(x-5)(x+3)}$

$=dfrac{2(x+3)-1}{(x-5)(x+3)}$

$=dfrac{2x+6-1}{(x-5)(x+3)}$

$$
=dfrac{2x+5}{(x-5)(x+3)}
$$

$dfrac{x^2-x-2}{4x^2-7x-2}divdfrac{x^2-2x-3}{3x^2-8x-3}$

$=dfrac{x^2+x-2x-2}{4x^2-8x+x-2}cdotdfrac{3x^2-9x+x-3}{x^2+x-3x-3}$

$=dfrac{x(x+1)-2(x+1)}{4x(x-2)+(x-2)}cdotdfrac{3x(x-3)+(x-3)}{x(x+1)-3(x+1)}$

$=dfrac{(cancel{x+1})(cancel{x-2})}{(cancel{x-2})(4x+1)}cdotdfrac{(cancel{x-3})(3x+1)}{(cancel{x+1})(cancel{x-3})}$

$$
=dfrac{3x+1}{4x+1}
$$

a) $dfrac{x-3}{2x+1}$; b) $dfrac{2x-7}{x-9}$; c) $5$; d) $dfrac{x+2}{2x+3}$; e) $dfrac{2x+5}{(x-5)(x+3)}$; f) $dfrac{3x+1}{4x+1}$

Let’s consider $dfrac{p(x)}{q(x)},dfrac{a(x)}{b(x)}$, where $b(x)not=0, q(x)not=0$.

$dfrac{p(x)}{q(x)}+dfrac{a(x)}{b(x)}=dfrac{p(x)b(x)+a(x)q(x)}{b(x)q(x)}$

$dfrac{p(x)}{q(x)}-dfrac{a(x)}{b(x)}=dfrac{p(x)b(x)-a(x)q(x)}{b(x)q(x)}$

$dfrac{p(x)}{q(x)}cdotdfrac{a(x)}{b(x)}=dfrac{p(x)a(x)}{q(x)b(x)}$

$dfrac{p(x)}{q(x)}divdfrac{a(x)}{b(x)}=dfrac{p(x)}{q(x)}cdotdfrac{b(x)}{a(x)}=dfrac{p(x)b(x)}{q(x)a(x)}$

The addition,subtraction and multiplication of two rational expressions lead to a rational expression, therefore the rational expressions are a closed set under these 3 operations.

In case of division, if $a(x)=0$ the result of division is undefined, therefore the rational expressions are not a closed set under division.

$$
dfrac{2x}{3x^2+16x+5}+dfrac{10}{3x^2+16x+5}
$$

$dfrac{2x}{3x^2+16x+5}+dfrac{10}{3x^2+16x+5}=dfrac{2x+10}{3x^2+16x+5}$

$=dfrac{2(x+5)}{3x^2+15x+x+5}$

$=dfrac{2(x+5)}{3x(x+5)+(x+5)}$

$=dfrac{2(cancel{x+5})}{(cancel{x+5})(3x+1)}$

$$
=dfrac{2}{3x+1}
$$

$$
dfrac{x^2-x-12}{3x^2-11x-4}cdotdfrac{3x^2-20x-7}{x^2-9}
$$

$dfrac{x^2-x-12}{3x^2-11x-4}cdotdfrac{3x^2-20x-7}{x^2-9}$

$=dfrac{x^2+3x-4x-12}{3x^2-12x+x-4}cdotdfrac{3x^2-21x+x-7}{(x-3)(x+3)}$

$=dfrac{x(x+3)-4(x+3)}{3x(x-4)+(x-4)}cdotdfrac{3x(x-7)+(x-7)}{(x-3)(x+3)}$

$=dfrac{(cancel{x+3})(cancel{x-4})}{(cancel{x-4})(cancel{3x+1})}cdotdfrac{(x-7)(cancel{3x+1})}{(x-3)(cancel{x+3})}$

$$
=dfrac{x-7}{x-3}
$$

$$
dfrac{2x^2+8x-10}{2x^2+15x+25}divdfrac{4x^2+20x-24}{2x^2+x-10}
$$

$dfrac{2x^2+8x-10}{2x^2+15x+25}divdfrac{4x^2+20x-24}{2x^2+x-10}$

$=dfrac{2x^2-2x+10x-10}{2x^2+10x+5x+25}divdfrac{(4x^2+20x+25)-49}{2x^2-4x+5x-10}$

$=dfrac{2x(x-1)+10(x-1)}{2x(x+5)+5(x+5)}divdfrac{(2x+5)^2-49}{2x(x-2)+5(x-2)}$

$=dfrac{2(x-1)(x+5)}{(x+5)(2x+5)}cdot dfrac{(x-2)(2x+5)}{(2x+5-7)(2x+5+7)}$

$=dfrac{2(x-1)(x+5)}{(x+5)(2x+5)}cdot dfrac{(x-2)(2x+5)}{(2x-2)(2x+12)}$

$=dfrac{cancel{2}(cancel{x-1})(cancel{x+5})}{(cancel{x+5})(cancel{2x+5})}cdot dfrac{(x-2)(cancel{2x+5})}{cancel{2}(cancel{x-1})cdot 2(x+6)}$

$$
=dfrac{x-2}{2(x+6)}
$$

$$
dfrac{7}{x+5}-dfrac{4-6x}{x^2+10x+25}
$$

$dfrac{7}{x+5}-dfrac{4-6x}{x^2+10x+25}$

$=dfrac{7}{x+5}-dfrac{4-6x}{(x+5)^2}$

$=dfrac{7}{x+5}cdotdfrac{x+5}{x+5}-dfrac{4-6x}{(x+5)^2}$

$=dfrac{7(x+5)-4+6x}{(x+5)^2}$

$=dfrac{7x+35-4+6x}{(x+5)^2}$

$$
=dfrac{13x+31}{(x+5)^2}
$$

a) $dfrac{2}{3x+1}$; b) $dfrac{x-7}{x-3}$; c) $dfrac{x-2}{2(x+6)}$; d) $dfrac{13x+31}{(x+5)^2}$;

$P=1000$

$r=0.24$

$$
n=1
$$

$A=Pleft(1+dfrac{r}{n}right)^{nt}$

$A=1000(1+0.24)^{1994-1865}$

$A=1000(1.24)^{129}$

$$
A=1.125634cdot 10^{15}
$$

b) We determine the amount of money that should be cashed after $1994-1865$ years:

c) The first digit in the $1.125634cdot 10^{15}$ is in the quadrillion place.

$A_1=Pe^{rt}$

$A_1=1000e^{0.24cdot 129}$

$$
A_1approx 2.790983cdot 10^{16}
$$

$A_1-A=2.790983cdot 10^{16}-1.125634cdot 10^{15}$

$=2.790983cdot 10^{16}-0.1125634cdot 10^{15}$

$$
=2.6784196cdot 10^{16}
$$

$x_1=3$

$x_2=1-i$

$$
x_3=1+i
$$

$f(x)=a(x-x_1)(x-x_2)(x-x_3)$

$f(x)=a(x-3)(x-1+i)(x-1-i)$

$f(x)=a(x-3)[(x-1)^2-i^2]$

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

$$
f(x)=a(x-3)(x^2-2x+2)
$$

$a=1$

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

$f(x)=x^3-2x^2+2x-3x^2+6x-6$

$$
f(x)=x^3-5x^2+8x-6
$$

$$
f(x)=x^3-5x^2+8x-6
$$

$$
F=180text{textdegree}-44text{textdegree}-58text{textdegree}=78text{textdegree}
$$

We determine the angle $F$:

$dfrac{sin F}{AB}=dfrac{sin A}{BF}$

$BFsin F=ABsin A$

$BF=dfrac{8cdot sin 44text{textdegree}}{sin 78text{textdegree}}$

$=dfrac{8cdot 0.69465837}{0.9781476}$

$$
approx 5.7
$$

$5.7$ miles

$$
sin (theta)=dfrac{1}{2},0text{textdegree}leq theta<360text{textdegree}
$$

$theta_1=30text{textdegree}$

$theta_2=150text{textdegree}$

$$
tan (theta)=sqrt 3,0text{textdegree}leq theta<360text{textdegree}
$$

$theta_1=60text{textdegree}$

$$
theta_2=240text{textdegree}
$$

$$
cos (theta)=dfrac{sqrt 3}{2},0text{textdegree}leq theta<360text{textdegree}
$$

$theta_1=30text{textdegree}$

$$
theta_2=330text{textdegree}
$$

$$
sin (theta)=-dfrac{sqrt 2}{2},0text{textdegree}leq theta<360text{textdegree}
$$

$theta_1=225text{textdegree}$

$$
theta_2=315text{textdegree}
$$

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

We start from the parent function $f_0(x)=x^2$. We shift $f_0$ 2 units to the right to get $f_1(x)=(x-2)^2$, then we vertically stretch $f_1$ by a factor of 2 to get $f_2(x)=2(x-2)^2$, then we reflect $f_2$ across the $x$-axis to get $f_3(x)=-2(x-2)^2$ and finally we shift $f_3$ 3 units up to get $f(x)$:

$$
f(x)=(x-1)^3+3
$$

We start from the parent function $f_0(x)=x^3$. We shift $f_0$ 1 unit to the right to get $f_1(x)=(x-1)^3$, then we shift $f_1$ 3 units up to get $f(x)$:

In this sum, we can notice that first and last terms, the constant difference between two terms are the following:

$$
a=-10,text{ } l=40,text{ } d=-9-(-10)=-9+10=1
$$

Now, we will calculate the number of terms in this sum:

$$
n=dfrac{l-a}{d}+1=dfrac{40-(-10)}{1}+1=51
$$

Finally, the required sum is the following:

$$
S=dfrac{n}{2}(a+l)=dfrac{51}{2}(-10+40)=dfrac{51}{2}cdot 30=51cdot 15=765
$$

$S=765$

$$
x=5
$$

If we consider the 2 dimensions coordinate system, the equation $x=5$ represents a vertical line, 5 units from the $y$-axis:

In the 3 dimensions coordinate system, the equation $x=5$ represents a plane parallel to the $yOz$ plane and 5 units above it:

$$
x+2y=5
$$

$$
x+2y+z=5
$$

$x=0, y=0Rightarrow z=5$

$x=0, z=0Rightarrow y=2.5$

$y=0, z=0Rightarrow x=5$

a) The point in which the $x$-axis and the $y$-axis intersect is the origin of the three-dimensional coordinate system. The value of $z$ at this point is zero.

$$
(0,0,0)
$$

b) The coordinates for the point described in part $(a)$ are:

$$
(4,2,3)
$$

a) The coordinates of the point for which $x=4$, $y=2$ and $z=3$ (as we lift the marker 3 units up from the plane $xOy$) are:

$$
(3,4,2)
$$

b) The coordinates of the point for which $x=3$, $y=4$ and $z=2$ are:

$$
xgeq 0, ygeq 0, zgeq 0
$$

$xgeq 0, ygeq 0, zleq 0$

$xgeq 0, yleq 0, zgeq 0$

$xgeq 0, yleq 0, zleq 0$

$xleq 0, ygeq 0, zgeq 0$

$xleq 0, ygeq 0, zleq 0$

$xleq 0, yleq 0, zgeq 0$

$xleq 0, yleq 0, zleq 0$

Thus all in all we need 8 regions (called $textcolor{#4257b2}{octants}$):

a) We build a $2times 2times 2$ rectangular prism:

b) We build a rectangular prism which is 2 units in length along the $x$-axis, 1 unit in length along the $y$-axis and 3 units in length along the $z$-axis,:

c) We label the coordinates of all the vertices for the prism in pat $(b)$:

We draw a rectangular prism that has vertices at $(1,0,0), (0,3,0), (0,0,4)$ (for a bigger drawing we considered the distance between 2 dots as being 0.5!):

b) We move the prism so that the vertices are at $(-1,0,0),(0,3,0), (0,0,4)$:

c) The farthest vertex from the origin in part $(b)$ is $(-1,3,4)$. We can build a prism like $OCMNQPD’A’$:

$A(0,1,-1)$

$B(1,2,0)$

$C(2,3,1)$

a) We plot the points (we consider the distance between 2 dots as 0.5$):

From the drawing the 3 points seem to be collinear in the plane $xOy$.

b) We should use different units of measure on the 3 axis: for example the distance between 2 dots to be 1 unit on the $x$-axis, $dfrac{1}{2}$ units on the $y$-axis and $dfrac{1}{3}$ units on the $z$-axis:

$B(0,1,2)$

$$
C(2,2,4)
$$

c) We identify the coordinates of 2 points that appear to be the same as $A(-2,0,0)$:

$$
textrm {Plotting Points in xyz-Space}
$$

We can graph points in three dimensions by detecting the three coordinates of the x-, y- and z-axis and drawing the point with these coordinates of the 3-dimension coordinate grid.

The following is the sample 3 dimension graph containing three points:

For example the point $B=(3, 4, 2)$ represents $x=3$, $y=4$ and $z=2$

begin{center}
begin{tabular}{|| c|c|c|c| c||}
hline
& Points on the & Points on the & Points on the & Points not \
& $x$-axis & $y$-axis & $z$-axis & on the axis \[0.5ex]
hline
1st point & $(1,0,0)$ & $(0,2,0)$ & $(0,0,1)$ & $(1,-2,1)$ \
hline
2nd point & $(-2,0,0)$ & $(0,1,0)$ & $(0,0,-1)$ & $(-3,1,2)$ \
hline
3rd point & $(2,0,0)$ & $(0,-1,0)$ & $(0,0,3)$ & $(-1,5,3)$\
hline
4th point & $(3,0,0)$ & $(0,-3,0)$ & $(0,0,-2)$ & $(-2,4,1)$\[1ex]
hline
end{tabular}
end{center}

b) The coordinates of the points situated on the $x$-axis have the general form $(x,0,0)$, the coordinates of the points situated on the $y$-axis have the general form $(0,y,0)$, the coordinates of the points situated on the $z$-axis have the general form $(0,0,z)$.

The coordinates of the points situated on one of the axis are zero for the positions corresponding to the other two axis.

(a) Factoring gives; $dfrac{4x}{(x+2)(x-4)}+dfrac{4}{x-4}timesdfrac{x+4}{x+4}$

$$
=dfrac{4x+4x+16}{(x+2)(x-4)}=dfrac{8(x+2)}{(x+2)(x-4)}
$$

$$
=dfrac{8}{(x-4)}
$$

(b) Factoring gives; $dfrac{16x-12}{(4x-3)(x+2)}-dfrac{3}{x+2}timesdfrac{4x-3}{4x-3}$

$$
=dfrac{4x-12-12x+9}{(4x-3)(x+2)}=dfrac{4x-3}{(4x-3)(x+2)}
$$

$$
=dfrac{1}{(x+2)}
$$

(a) $dfrac{8}{(x-4)}$ (b) $dfrac{1}{(x+2)}$

a-

$dfrac{(x-3)^2}{2x-1}cdot dfrac{2x-1}{(3x-14)(x+6)}cdot dfrac{x+6}{x-3}$          (Given)

$=dfrac{(x-3)(x-3)(2x-1)(x+6)}{(2x-1)(3x-14)(x+6)(x-3)}$

$=dfrac{ cancel {(x-3)} (x-3) cancel {(2x-1)} cancel {(x+6)}}{ cancel {(2x-1)} (3x-14) cancel {(x+6)} cancel {(x-3)}}$          (Cancel the common factors)

$=dfrac {x-3}{3x-14}$          (Simplify)

b-

$dfrac{(4x^2+5x-6)}{(3x^2+5x-2)}div dfrac{4x^2+x-3}{6x^2-5x+1}$          (Given)

$dfrac{(4x^2+5x-6)}{(3x^2+5x-2)} times dfrac {6x^2-5x+1}{4x^2+x-3}$          (Multiply by receprocal)

$=dfrac{(4x^2+5x-6)(6x^2-5x+1)}{(3x^2+5x-2)(4x^2+x-3)}$

$=dfrac{(x+2)(4x-3)(3x-1)(2x-1)}{(x+2)(x+1)(3x-1)(4x-3)}$          (Factor polynomials)

$=dfrac{ cancel {(x+2)} cancel {(4x-3)} cancel {(3x-1)} (2x-1)}{cancel {(x+2)} (x+1) cancel {(3x-1)} cancel {(4x-3)}}$          (Cancel common factors)

$=dfrac {2x-1}{x+1}$          (Simplify)

a-          $dfrac{(x-3)^2}{2x-1}cdot dfrac{2x-1}{(3x-14)(x+6)}cdot dfrac{x+6}{x-3}=dfrac {x-3}{3x-14}$

b-          $dfrac{(4x^2+5x-6)}{(3x^2+5x-2)}div dfrac{4x^2+x-3}{6x^2-5x+1}=dfrac {2x-1}{x+1}$

$$
(a+b)^4
$$

$(a+b)^4=a^4+_4C_1a^3b+_4C_2a^2b^2+_4C_3ab^3+b^4$

$=a^4+dfrac{4!}{3!1!}a^3b+dfrac{4!}{2!2!}a^2b^2+dfrac{4!}{1!3!}ab^3+b^4$

$=a^4+dfrac{3!cdot 4}{3!1!}a^3b+dfrac{2!cdot 3cdot 4}{2!cdot 1cdot 2}a^2b^2+dfrac{3!cdot 4}{1cdot 3!}ab^3+b^4$

$$
=a^4+4a^3b+6a^2b^2+4ab^3+b^4
$$

We use the Binomial Theorem:

$(a+b)^n=_nC_0a^n+_nC_1a^{n-1}b+….+_nC_{n-1}ab^{n-1}+_nC_nb^n$.

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

$(3m-2)^4$

$=(3m)^4+_4C_1(3m)^3(-2)+_4C_2(3m)^2(-2)^2+_4C_3(3m)(-2)^3+(-2)^4$

$=81m^4+dfrac{4!}{3!1!}(27m^3(-2)+dfrac{4!}{2!2!}(9m^2)(-2)^2-dfrac{4!}{1!3!}(3m)(-2)^3+(-2)^4$

$=81m^4-dfrac{3!cdot 4}{3!1!}(54m^3)+dfrac{2!cdot 3cdot 4}{2!cdot 1cdot 2}(36m^2)-dfrac{3!cdot 4}{1cdot 3!}(24m)+16$

$$
=81m^4-216m^3+216m^2-96m+16
$$

We use the Binomial Theorem:

$(a+b)^n=_nC_0a^n+_nC_1a^{n-1}b+….+_nC_{n-1}ab^{n-1}+_nC_nb^n$.

a) $a^4+4a^3b+6a^2b^2+4ab^3+b^4$

b) $81m^4-216m^3+216m^2-96m+16$

Given ,

$$
begin{align*}
f (x) & = 1 + tan(x – dfrac{pi}{4})\
end{align*}
$$

In order to sketch the graph of $f (x)$, we will first assume the given function as $f_{o} ( x) = tan (x)$ for better understanding of the graph. We will call this assumption a parent function.

As we can clearly see that the graph of $f( x)$ is deriving from the assumed parent function by shifting the parent function by $dfrac{pi}{4}$ units to the right and one unit up.

The opposite figure represents the post $(overline {AC})$ and its shadow $(overline {AB})$

$tan B=dfrac {3.5}{4.25}=0.8235$

$mangle B=39.47$ $text{textdegree}$

The opposite figure represents the tree $(overline {DF})$ which tall is $x$ feet and its shadow $(overline {DE})$ which tall is 100 feet

$mangle E=39.47$ $text{textdegree}$

$tan E=dfrac {x}{100}=0.8235$

$x=tan E cdot 100$

$x=0.8235 cdot 100$

$x=82.35$ feet

$$
h(t)=80t-16t^2
$$

a) We graph the function $h(t)$ for object A:

$h(t)=0$

$80t-16t^2=0$

$16t(5-t)=0$

$t_1=0$

$$
5-t=0Rightarrow t_2=5
$$

We determine the $x$-intercepts of the function $h$:

$$
t_2-t_1=5-0=5
$$

As object B stays in the air around $5.7$ seconds, it means object B stays in air longer.

$$
h_B=68
$$

b) From the graph for object B we find the maximum height $h_B$:

$h(t)=-16(t^2-5t)=-16left(y^2-5t+dfrac{25}{4}right)+16cdot dfrac{25}{4}$

$$
=-16(t-2.5)^2+100
$$

We write the function $h(t)$ in vertex form:

$$
(2.5,100)
$$

$h(t)>64$

$80t-16t^2>64$

$16t^2-80t+64<0$

$16(t^2-5t+4)<0$

$16(t^2-4t-t+4)<0$

$16[t(t-4)-(t-4)]<0$

$16(t-4)(t-1)<0$

$(t-4)(t-1)<0$

$$
tin (1,4)
$$

$$
[0,5]
$$

d) The domain for function $h$ is:

a-

The graph of the equation $5x+8y+10z=40$ is a plane in the 3-dimensional coordinate grid.

b-

To know if the point $(4, 5, -2)$ a solution to the equation or not, we substitute with the point in the equation.

$5(4)+8(5)+10(-2)=20+40-20=40$

So, the point $(4, 5, -2)$ is a solution, and this clear from the graph as the point lies on the plane of the equation.

a-          The shape formed by the equation is a plane.

b-          The point $(4, 5, -2)$ is a solution for the equation.

We can graph an equation like $12x+4y+5z=60$ in three dimensions, by graphing x-, y- and z-intercepts and connecting the three intercepts, then extending the resulting plane.

The following is the graph of the equation.

a-

The graph of the equation $5x+8y=40$ is a plane.

We can graph an equation like $12x+4y+5z=60$ in three dimensions, by graphing x-, y- and z-intercepts and connecting the three intercepts, then extending the resulting plane.

a-

The graph of the equation $12x+4y+5z=60$.

It is plane in the 3D coordinates graph.

$$
3x+4y+2z=12
$$

$x=0,y=0Rightarrow 3(0)+4(0)+2z=12Rightarrow 2z=12Rightarrow z=6Rightarrow A(0,0,6)$

$y=0,z=0Rightarrow 3x+4(0)+2(0)=12Rightarrow 3x=12Rightarrow x=4Rightarrow B(4,0,0)$

$x=0,z=0Rightarrow 3(0)+4y+2(0)=12Rightarrow 4y=12Rightarrow y=3Rightarrow C(0,3,0)$

We graph the plane determined by the points $A, B, C$:

$$
12x-9y+108=0
$$

$x=0,z=0Rightarrow 12(0)-9y+108=0Rightarrow 9y=108Rightarrow y=12Rightarrow A(0,12,0)$

$y=0,z=0Rightarrow 12x-9(0)+108=0Rightarrow 12x=-108Rightarrow x=-9Rightarrow B(-9,0,0)$

We graph the plane determined by the points $A, B$ and parallel to the $z$-axis:

$$
x=4
$$

c) We graph the equation in three dimensions (a plane parallel to the plane $yOz$):

$6y+15(0)-60=0$ ; $y=10$

(a) Get the y and z intercepts by setting either term to zero and solving. The x-intercept is zero here, with the function restricted to the y-z plane $;$

$15z+6(0)-60=0$ ; $z=4$

so we have, in ordered triple notation $left( 0,10,4right)$

$x=dfrac{24}{3}=8$ ; $y=dfrac{24}{4}=6$ ; $z=dfrac{24}{2}=12$

(b) $(8,6,12)$

$(x+3)^{2}=25$ ; $x=2$ ; $z=4$ $z^{2}=25-9$ ; $z=4$

(c) $(2,0,4)$ Note we can check this is correct by observing that we have an equation of a circle with centre on the x-axis at x=-3 and of radius 5. (-3+5)=2

(d) $(0,0,6)$

a-

$dfrac {1}{x+2}+dfrac {3}{x^2-4}$          (Given)

$dfrac {1}{x+2}+dfrac {3}{(x+2)(x-2)}$          (Factoring $x^2-4$ by the difference of two squares)

$dfrac {1}{x+2} cdot dfrac {x-2}{x-2}+dfrac {3}{(x+2)(x-2)}$          ($(x+2)(x-2)$ is the least common multiple)

$=dfrac {x-2+3}{(x+2)(x-2)}$

$=dfrac {x+1}{(x+2)(x-2)}$

b-

$dfrac {3}{2x+4} – dfrac {x}{x^2+4x+4}$          (Given)

$dfrac {3}{2(x+2)}-dfrac {x}{(x+2)^2}$          (Factoring)

The least common multiple is $2(x+2)^2$

$dfrac {3}{2(x+2)} cdot dfrac {x+2}{x+2}-dfrac {2}{2}dfrac {x}{(x+2)^2}$

$dfrac {3x+6}{2(x+2)^2} – dfrac {2x}{2(x+2)^2}$

$dfrac {3x+6-2x}{2(x+2)^2}$

$dfrac {x+6}{2(x+2)^2}$

c-

$dfrac {x^2+5x+6}{x^2-9} cdot dfrac {x-3}{x^2-2x}$          (Given)

$dfrac {(x+3)(x+2)}{(x+3)(x-3)} cdot dfrac {x-3}{x(x+2)}$          (Factoring numerators and denominators)

$dfrac {cancel {(x+3)} cancel {(x+2)}}{ cancel {(x+3)} cancel {(x-3)}} cdot dfrac { cancel {(x-3)}}{x cancel {(x+2)}}$          (Common factors)

$$
=dfrac {1}{x}
$$

d-

$dfrac {4}{x-2} div dfrac {8}{2-x}$          (Given)

$dfrac {4}{x-2} times dfrac {-(x+2)}{8}$          (Multiply by reciprocal)

$dfrac { cancel {4}}{ cancel {(x-2)}} times dfrac {-1 cdot cancel {(x-2)}}{2 cdot cancel {4}}$

$=dfrac {-1}{2}$

a-          $dfrac {1}{x+2}+dfrac {3}{x^2-4}=dfrac {x+1}{(x+2)(x-2)}$

b-          $dfrac {3}{2x+4} – dfrac {x}{x^2+4x+4}=dfrac {x+6}{2(x+2)^2}$

c-          $dfrac {x^2+5x+6}{x^2-9} cdot dfrac {x-3}{x^2-2x}=dfrac {1}{x}$

d-          $dfrac {4}{x-2} div dfrac {8}{2-x}=dfrac {-1}{2}$

$$
f(x)=a(x-3)^3+3
$$

The given graph is the graph of the parent function $y=x^3$, shifted 3 units to the right, vertically stretched and 3 units up:

$a(5-3)^3+3=4$

$8a+3=4$

$8a=1$

$a=dfrac{1}{8}$

We determine $a$ using the point $(5,4)$:

$$
f(x)=dfrac{1}{8}(x-3)^3+3
$$

$y=dfrac{1}{8}(x-3)^3+3$

$x=dfrac{1}{8}(y-3)^3+3$

$x-3=dfrac{1}{8}(y-3)^3$

$8(x-3)=(y-3)^3$

$2sqrt[3]{x-3}=y-3$

$y=2sqrt[3]{x-3}+3$

$$
f^{-1}(x)=2sqrt[3]{x-3}+3
$$

$90text{textdegree}<A<180text{textdegree}$

$$
sin A=dfrac{3}{10}
$$

$sin^2 A+cos^2 A=1$

$cos^2 A=1-left(dfrac{3}{10}right)^2$

$=1-dfrac{9}{100}=dfrac{91}{100}$

$$
cos A=pmdfrac{sqrt{91}}{10}
$$

$$
cos A=-dfrac{sqrt{91}}{10}
$$

$$
tan A=dfrac{sin A}{cos A}=dfrac{dfrac{3}{10}}{-dfrac{sqrt{91}}{10}}=-dfrac{3}{sqrt{91}}=-dfrac{3sqrt{91}}{91}
$$

$$
-dfrac{3sqrt{91}}{91}
$$

$sin 60text{textdegree}=dfrac{h}{12}$

$$
h=12sin 60text{textdegree}=12cdot dfrac{sqrt 3}{2}=6sqrt 3
$$

$$
V=7cdot 13cdot 6sqrt 3=546sqrt 3
$$

$dfrac{25}{546sqrt 3}approx 0.026$ fish/cubic inch

$dfrac{25}{dfrac{546sqrt 3}{12^3}}=approx 45.68$ fish/cubic foot

$$
y=2x^2+7x-7
$$

$y=2left(x^2+dfrac{7}{2}xright)-7$

$=2left(x^2+dfrac{7}{2}x+dfrac{49}{16}right)-2cdotdfrac{49}{16}-7$

$$
=2left(x+dfrac{7}{4}right)^2-dfrac{105}{8}
$$

$$
left(-dfrac{7}{4},-dfrac{105}{8}right)
$$

$$
x=-dfrac{7}{4}
$$

$$
y=3x^2-x-8
$$

$y=3left(x^2-dfrac{1}{3}xright)-8$

$=3left(x^2-dfrac{1}{3}x+dfrac{1}{36}right)-3cdotdfrac{1}{36}-8$

$$
=3left(x-dfrac{1}{6}right)^2-dfrac{97}{12}
$$

$$
left(dfrac{1}{6},-dfrac{97}{12}right)
$$

$$
x=dfrac{1}{6}
$$

$12x-2y=16 rightarrow 2y=12x-16 rightarrow y=6x-8$          (1)

$30x+2y=68 rightarrow 2y=68-30x rightarrow y=34-15x$          (2)

$6x-8=34-15x$          (Equating $y$ from both equations.)

$6x+15x=34+8$

$21x=42$

$$
x=2
$$

$y=6(2)-8$          (Substituting 4 for $x$ in equation (1))

$$
y=4
$$

The solution of the system is: $(2, 4)$

The solution of the system is: $(2, 4)$

$$
begin{cases}
-2x+3y+4z=1\
6x-3y+4z=11\
3x-3y+2z=10
end{cases}
$$

$y=0,z=0Rightarrow -2x=1Rightarrow x=-dfrac{1}{2}$

$z=0,x=0Rightarrow 3y=1Rightarrow y=dfrac{1}{3}$

$x=0,y=0Rightarrow 4z=1Rightarrow z=dfrac{1}{4}$

$y=0,z=0Rightarrow 6x=11Rightarrow x=dfrac{11}{6}$

$z=0,x=0Rightarrow -3y=11Rightarrow y=-dfrac{11}{3}$

$x=0,y=0Rightarrow 4z=11Rightarrow z=dfrac{11}{4}$

$y=0,z=0Rightarrow 3x=10Rightarrow x=dfrac{10}{3}$

$z=0,x=0Rightarrow -3y=10Rightarrow y=-dfrac{10}{3}$

$x=0,y=0Rightarrow 2z=10Rightarrow z=5$

For $textcolor{#4257b2}{substitution}$, we choose one of the variables and isolate in one of the equation. We will express it in terms of the other two variables. We leave this equation aside and replace that variable in the other two equations, which will lead us to a system of two equations with two variables. We solve this smaller system and get back to the substitution we made in order to find the third variable.

For $textcolor{#4257b2}{elimination}$, we multiply and add equations so that we eliminate one of the variables.We will get a smaller system with two equations and two variables. We solve it, then we get back to any of the original equations and determine the third variable.

For $textcolor{#4257b2}{Cramer}$, we need to work with determinants.

$$
begin{cases}
-2x+3y+4z=1\
6x-3y+4z=11\
3x-3y+2z=10
end{cases}
$$

$-2x+3y+4z+6x-3y+4z=1+11$

$$
4x+8z=12
$$

$-(3x-3y+2z)+6x-3y+4z=-10+11$

$-3x+3y-2z+6x-3y+4z=1$

$$
3x+2z=1
$$

$$
begin{cases}
4x+8z=12\
3x+2z=1
end{cases}
$$

$$
begin{cases}
x+2z=3\
3x+2z=1
end{cases}
$$

$$
begin{cases}
-x-2z=-3\
3x+2z=1
end{cases}
$$

$-x-2z+3x+2z=-3+1$

$2x=-2$

$textcolor{#4257b2}{x=-1}$

$x+2z=3$

$2z=3-x=3-(-1)=4$

$$
textcolor{#4257b2}{z=2}
$$

$-2x+3y+4z=1$

$-2(-1)+3y+4(2)=1$

$2+3y+8=1$

$3y=-9$

$$
textcolor{#4257b2}{y=-3}
$$

c) We stlll need to solve for $y$. For this we replace the values of $x$ and $z$ in one of the original system’s equations:

$$
(x,y,z)=(-1,-3,2)
$$

#### (a)

In order to solve the given system of equations, we will express one of three variables in terms of the other.

For example, we will express $x$ from the first equation and substitute it in other two equations:

$$
x+y+3z=3Rightarrow x=3-3z-y
$$

$$
Rightarrow 2x+y+6z=2(3-3z-y)+y+6z=6-6z-2y+y+6z=6-y=2Rightarrow y=4
$$

$$
Rightarrow 2x-y+3z=2(3-3z-y)-y+3z=6-6z-2y-y+3z=6-3z-3y=-7, y=4
$$

$$
Rightarrow 6-3z-12=-7Rightarrow -6-3z=-7Rightarrow 3z=1Rightarrow z=dfrac{1}{3}
$$

$$
Rightarrow x=3-3dfrac{1}{3}-4=3-1-4=-2
$$

So, the solution is $(x,y,z)=(-2,4,tfrac{1}{3})$.

#### (b)

Here, we can notice that we actually have two equations because the first two equations are the same.

But, we have three variables, so, the conclusion is that we can not solve this system.

#### (c)

In order to solve the given system of equations, we will express one of three variables in terms of the other.

For example, we will express $z$ from the second equation and substitute it in other two equations:

$$
-2x+2y+z=5Rightarrow z=5+2x-2y
$$

$$
Rightarrow 5x-4y-6(5+2x-2y)=5x-4y-30-12x+12y=16y-7x-30=-19
$$

$$
Rightarrow 16y-7x=11
$$

$$
Rightarrow 3x-6y-5(5+2x-2y)=3x-6y-25-10x+10y=4y-7x-25=-16
$$

$$
Rightarrow 4y-7x=9
$$

Now, we need to solve the following system:

$$
16y-7x=11
$$

$$
4y-7x=9
$$

In order to solve the previous system, we will subtract them and get:

$$
16y-7x-4y+7x=11-9Rightarrow 12y=2Rightarrow y=6
$$

Now, we will substitute $6$ for $y$ in one of the previous two equations
in order to get $x$:

$$
4cdot 6-7x=9Rightarrow 7x=15Rightarrow x=dfrac{15}{7}
$$

$$
Rightarrow z=5+2x-2y=5+2cdotdfrac{15}{7}-2cdot 6=5+dfrac{30}{7}-24=dfrac{35+30-168}{7}=-dfrac{103}{7}
$$

The final solution is $(x,y,z)=left(dfrac{15}{7}, 6, -dfrac{103}{7} right)$.

#### (d)

From the first equation, we get that

$$
6x+4y=12-z
$$

Now, we will substitute this in the second equation and get:

$$
12-z+2z=12Rightarrow 12+z=12Rightarrow z=0
$$

Because we got $z=0$, now we get only the following equation:

$$
6x+4y=12
$$

We have one equation but two variables, so, the conclusion is that we can not solve the given system.

a) $(x,y,z)=(-2,4,tfrac{1}{3})$; b) It can not be solved; c) $(x,y,z)=left(dfrac{15}{7}, 6, -dfrac{103}{7} right)$; d) It can not be solved.

We can describe the intersection as a line which lies in $xy$-plane.

$$
textrm {Systems ~of ~Three ~Equations ~with ~Three ~Variables}
$$

$$
begin{enumerate}
item For solving a system of three equations with three variables, We follow the following steps:\
item Solving each pair of the equations so that we eliminate one of the variables and get 2 equations with two variables.\
item Solving the two equations to get the values of the two variables.\
item Substituting in one of the original equations with the values of the two variables so that calculating the value of the third one.\
item Checking for the solution in the other two equations.\
end{enumerate}
$$

$$
begin{cases}
10x+6y+5z=30\
6x+15y+5z=30
end{cases}
$$

$y=0,z=0Rightarrow 10x=30Rightarrow x=3$

$z=0,x=0Rightarrow 6y=30Rightarrow y=5$

$x=0,y=0Rightarrow 5z=30Rightarrow z=6$

$y=0,z=0Rightarrow 6x=30Rightarrow x=5$

$z=0,x=0Rightarrow 15y=30Rightarrow y=2$

$x=0,y=0Rightarrow 5z=30Rightarrow z=6$

a)

$$
frac{2x^3 + 5x^2 -3x}{4x^-4x^2+x} = frac{x(2x^2+5x-3)}{x(4x^2-4x+1)}
$$

$$
=frac{2x^2+5x-3}{4x^2-4x+1} = frac{2x^2 +6x -x -3}{4x^2-2x-2x+1}
$$

$$
=frac{2x(x+3)-(x+3)}{2x(2x-1)-(2x-1)} = frac{(2x-1)(x+3)}{(2x-1)(2x-1)}
$$

$$
=frac{x+3}{2x-1}
$$

b)

$$
frac{3x^2-5x-2}{2x^2-11x+15}cdotfrac{2x^2-5x}{3x^3-5x^2-2x}
$$

$$
=frac{3x^2-6x+x-2}{2x^2-6x-5x+15}cdotfrac{x(2x -5)}{x(3x^2-5x-2)}
$$

$$
=frac{3x(x-2)+(x-2)}{2x(x-3)-5(x-3))}cdotfrac{2x-5}{3x(x-2)+(x-2)}
$$

$$
=frac{(3x+1)(x-2)}{(2x-5)(x-3)}cdotfrac{(2x-5)}{(3x+1)(x-2)}
$$

$$
=frac{1}{x-3}
$$

a) $frac{x+3}{2x-1}$ b) $frac{1}{x-3}$

$$
(a+b)^9
$$

$n=9$

$k=4$

$T_5=_9C_4a^{9-4}b^4=dfrac{9!}{5!4!}a^5b^4$

$=dfrac{cancel{5!}cdot 6cdot 7cdot 8cdot 9}{cancel{5!}cdot 1cdot 2cdot 3cdot 4}a^5b^4$

$$
=126a^5b^4
$$

In order to determine the fifth term we use the formula:

$T_{k+1}=_nC_k a^{n-k}b^k$.

$$
(2x+y)^8
$$

$n=8$

$k=4$

$T_5=_8C_4(2x)^{8-4}y^4=dfrac{8!}{4!4!}(2x)^4y^4$

$=dfrac{cancel{4!}cdot 5cdot 6cdot 7cdot 8}{cancel{4!}cdot 1cdot 2cdot 3cdot 4}(16x^4y^4)$

$$
=1120x^4y^4
$$

In order to determine the fifth term we use the formula:

$T_{k+1}=_nC_k a^{n-k}b^k$.

a) $126a^5b^4$

b) $1120x^4y^4$

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

$$
pm 1, pm 2, pm 3, pm 6
$$

$$
P(1)=1^3-2(1^2)-5(1)+6=0
$$

$polyhornerscheme[x=1]{x^3-2x^2-5x+6}$

$$
P(x)=(x-1)(x^2-x-6)
$$

We found that $x=1$ is a root. We divide $P(x)$ by $x-1$:

$P(x)=(x-2)(x^2+2x-3x-6)$

$=(x-2)[x(x+2)-3(x+2)]$

$$
=(x-1)(x+2)(x-3)
$$