Core Connections Integrated 3
Core Connections Integrated 3
1st Edition
Brian Hoey, Judy Kysh, Leslie Dietiker, Tom Sallee
ISBN: 9781603283939
Textbook solutions

All Solutions

Page 368: Closure Activity

Exercise 121
Step 1
1 of 4
Exercise scan
We are given the triangle:
Step 2
2 of 4
$x^2=28^2+25^2-2cdot 28cdot 25cdot cos 37text{textdegree}$

$=1409-1400cdot 0.79863551$

$=290.91019$

$$
x=sqrt{290.91019}approx 17.06
$$

Method 1: We use the Law of Cosines:
Step 3
3 of 4
$dfrac{sin 37text{textdegree}}{x}=dfrac{sin 81text{textdegree}}{28}$

$x=dfrac{sin 37text{textdegree}cdot 28}{sin 81text{textdegree}}$

$=dfrac{0.60181502cdot 28}{0.98768834}$

$$
approx 17.06
$$

Method 2: We use the Law of Sines:
Result
4 of 4
$$
x=17.06′
$$
Exercise 122
Step 1
1 of 5
a)

We have to write this equation in exponential form.

This means for equation $text{underline{$log_{b}x=y$}}$, exponential form will be $text{underline{$b^y=x$}}$.

$$
begin{align*}
log_{3}(2x-1)&=-2 & text{logarithimic equation}\
3^{-2}&=2x-1 & text{exponential form}\
dfrac{1}{9}+1&=2x-1+1 & text{Add $1$ to both sides}\
dfrac{10}{9}&=2x\
x&=dfrac{10}{2cdot9} & text{Divide both sides by $2$}\
x&=boxed{dfrac{5}{9}} implies & text{textbf{solution}}\
end{align*}
$$

Step 2
2 of 5
b)

Apply logarithimic rule $underline{log_{b}b^k=k}$

$$
begin{align*}
5^{log_{5}x}&=3\
x&=boxed{3} implies & text{textbf{solution}}
end{align*}
$$

Step 3
3 of 5
c)

First we will apply rule $underline{log_{b}a-log_{b}c=log_{b}dfrac{a}{c}}$

$$
begin{align*}
log_{2}(x)-log_{3}(x)&=4\
log_{2}dfrac{x}{3}&=4 & text{apply logarithimic rule}\
2^4&=dfrac{x}{3} & text{write in exponential form}\
dfrac{x}{3}&=16\
x&=boxed{48} & text{textbf{solution}}
end{align*}
$$

Step 4
4 of 5
d)

$log_{3}{5}$ can be calculated right away, so $x=log_{3}5approx1.465$

Result
5 of 5
a) $x=dfrac{5}{9}$

b) $x=3$

c) $x=48$

d) $xapprox1.465$

Exercise 123
Solution 1
Solution 2
Step 1
1 of 3
$$
begin{cases}
ygeq x^2\
ygeq (x-4)^2+2
end{cases}
$$
We are given the system:
Step 2
2 of 3
Exercise scan
We graph both parabolas and shade the region above them. The solution is the intersection of these regions:
Step 3
3 of 3
The graph of $ygeq (x-4)^2+2$ is positioned 4 units to the right and 2 units up in relation to the graph of $ygeq x^2$.
Step 1
1 of 2
Since both expressions are in the form $ax^2+bx+c$, graph will contain two parabolas.

Both parabolas are opening upwards since $a>0$ in both expressions.

Parabola of $x^2$ has vertex at point $(0,0)$ and parabola of $(x-4)^2+2$ has vertex $(4,2)$.

The graph of $(x-4)^2+2$ is graph of $x^2$ positioned $4$ units to the right and $2$ units up.

Step 2
2 of 2
Exercise scan
Exercise 124
Step 1
1 of 4
Interchange x and y in y=f(x) and solve for y where y becomes the inverse function $y^{-1}9x)$;
Step 2
2 of 4
$$
x=2+sqrt{2y-4}
$$
Step 3
3 of 4
$2y-4=(x-2)^{2}$ ; $2y=x^{2}-4x+4+4$
Step 4
4 of 4
$$
y^{-1}(x)=dfrac{x^{2}}{2}-2x+4
$$
Exercise 125
Step 1
1 of 5
We have the equation: $27=6^x$.
Solve the equation. Apply exponent rule. Apply $log$ on both sides.

$$
begin{align*}
ln left(27right)&=xln left(6right) \
x &= dfrac{ log 27}{log 6} \
x&approx 1.839
end{align*}
$$

Step 2
2 of 5
We have the equation: $27 = dfrac{1}{6^x}$.
Solve the equation. Apply exponent rules.

$$
begin{align*}
27 &= 6^{-x} \
log 27 &= log 6^{-x} \
x &= dfrac{log 27}{-log 6 } \
x &approx -1.839
end{align*}
$$

Step 3
3 of 5
We have the equation: $27 = dfrac{1}{left(dfrac{1}{6}right)^x}$.
Solve the equation. Apply exponent rules.

$$
begin{align*}
27 &= dfrac{1}{6^{-x}} \
27 &= 6^x \
&text{We solved this. (a)} \
x &approx 1.839
end{align*}
$$

Step 4
4 of 5
We have the equation: $dfrac{1}{left(dfrac{1}{6}right)^x} = 6^x$.
Solve the equation. Apply exponent rules.

$$
begin{align*}
dfrac{1}{left(dfrac{1}{6}right)^x} &= dfrac{1}{dfrac{1}{6^x}} \
&= 1 cdot dfrac{6^x}{1} \
&= 6^x
end{align*}
$$

All $x$.

Result
5 of 5
a) $1.839$
b) $-1.839$
c) $1.839$
d) All $x$
Exercise 126
Step 1
1 of 5
a)

By squaring both sides of the equation we get:

$$
begin{align*}
x+7&=(x+1)^2\
x+7&=x^2+2x+1 & text{ Use the fact $(x+1)^2=x^2+2x+1$}\
x+7-(x+7)&=x^2+2x+1-x-7 & text{Subtract $x+7$ from both sides}\
0&=x^2+x-6 & text{Combine like terms}\
x^2+3x-2x-6&=0 & text{Rewrite $x$ as $3x-2x$}\
x(x+3)-2x-6&=0 & text{ Factor out $x$ of first two terms}\
x(x+3)-2(x+3)&=0 & text{Factor out $2$ of second two terms}\
(x-2)(x+3)&= 0 & text{Factor out $x+3$}
end{align*}
$$

Now the solutions for equation $(x-2)(x+3)=0$ are $x=2$ or $x=-3$. We must check them.

Step 2
2 of 5
CHECKING FOR SOLUTION $X=2$

$$
begin{align*}
sqrt{x+7}&=x+1\
sqrt{2+7}&=2+1 & text{Substitute $2$ for $x$}\
sqrt{9}&=3\
3&=3 implies & text{The solution $x=2$ is correct.}\
end{align*}
$$

CHECKING FOR SOLUTION $X=-3$

$$
begin{align*}
sqrt{x+7}&=x+1\
sqrt{-3+7}&=-3+1 & text{Substitute $-3$ for $x$}\
sqrt{4}&=-2\
2&ne-2 implies & text{The solution is extraneous.}
end{align*}
$$

Step 3
3 of 5
b)

By subtracting $4$ from both sides of the equation we get:

$$
begin{align*}
3x^4&=77\
x^4&=dfrac{77}{3} & text{Divide both sides by $3$}\
sqrt{x^4}&=sqrt{dfrac{77}{3}} & text{Take square root of both sides}\
x^2&=sqrt{dfrac{77}{3}} & text{Use the fact $x^2>0$}\
x&=boxed{pm sqrt[4]{dfrac{77}{3}}} & text{Take square root of both sides}\
end{align*}
$$

Now we must check both solutions.

Step 4
4 of 5
CHECKING FOR SOLUTION $X=sqrt[4]{dfrac{77}{3}}$

$$
begin{align*}
3x^4+4&=81\
3left(sqrt[4]{dfrac{77}{3}}right)^4+4&=81\
3dfrac{77}{3}+4&=81\
77+4&=81\
81&=81 & text{The solution $x=sqrt[4]{dfrac{77}{3}}$ is correct}\
end{align*}
$$

CHECKING FOR SOLUTION $X=-sqrt[4]{dfrac{77}{3}}$

$$
begin{align*}
3x^4+4&=81\
3left(-sqrt[4]{dfrac{77}{3}}right)^4+4&=81\
3dfrac{77}{3}+4&=81\
77+4&=81\
81&=81 & text{The solution $x=-sqrt[4]{dfrac{77}{3}}$ is correct}\
end{align*}
$$

Result
5 of 5
a)$x=2$

b)$x=pm sqrt[4]{dfrac{77}{3}}$

Exercise 127
Step 1
1 of 6
$$
x(x-1)(2x+3)=0
$$
a) We are given the equation:
Step 2
2 of 6
$x=0Rightarrow x_1=0$

$x-1=0Rightarrow x_2=1$

$2x+3=0Rightarrow x_3=-dfrac{3}{2}$

Weuse the Zero Product Property to determine the solutions:
Step 3
3 of 6
$$
2x^3+x^2-3x=0
$$
b) We are given the equation:
Step 4
4 of 6
$x(2x^2+x-3)=0$

$x(2x^2-2x+3x-3)=0$

$x[2x(x-1)+3(x-1)]=0$

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

We factor and use the Zero Product Property:
Step 5
5 of 6
$$
left{-dfrac{3}{2}, 0, 1right}
$$
We got the sme equations as in part $(a)$. The solutions are:
Result
6 of 6
a) $left{-dfrac{3}{2}, 0, 1right}$

b) $left{-dfrac{3}{2}, 0, 1right}$

Exercise 128
Step 1
1 of 3
a-          An exponential function would best model the population over time.
Step 2
2 of 3
b-

Assuming $n$ is the number of years,          $a_1$ is the initial value, $r$ is the decreasing rate, and $a_n$ is the population after $n$ years.

$a_n=a_1 times (1-r)^n$

$70,379=72,000 times (1-r)^2$

$(1-r)^2=dfrac {70,379}{72,000}$

$(1-r)^2=0.9775$

$1-r=0.9887$

$r=0.0113=%1.13$

The equation that would model the changing population is:

$$
a_n=72,000 times (0.9887)^n
$$

For the population to level off eventually at 60,000:

$60,000=72,000 times (0.9887)^n$

$(0.9887)^n=dfrac {60}{72}$

$(0.9887)^n=dfrac {5}{6}$

$log (0.9887)^n=log (0.8333)$

$nlog (0.9887)=log (0.8333)$

$n cdot (-0.0049)=-0.0792$

$n=dfrac {-0.0792}{-0.0049}$

$$
n approx 16 mathrm { ~years}
$$

Result
3 of 3
a-          An exponential function would best model the population over time.

b-          The equation that would model the changing population is:

$$
a_n=72,000 times (0.9887)^n
$$

Exercise 129
Step 1
1 of 7
We need to calculate the percentiles for each person.
Step 2
2 of 7
Let $0$ be the lower bound and $6$ be the upper bound. We have an average of $4.2$ and a standard deviation of $1.2$.
Step 3
3 of 7
By using the normalcdf feature on the graphing calculator, we get the normal distribution of $0.93$. Therefore, the new walking distance of $6$ miles is in the $93$rd percentile.
Step 4
4 of 7
Now for April, we will take $0$ as the lower and $60$ as the upper bound. We have an average of $40$ and a standard deviation of $8$.
Step 5
5 of 7
By using the normalcdf feature, we get the normal distribution of $0.99$. Therefore, that is the $99$th percentile.
Step 6
6 of 7
Therefore, April has higher percentile so she is doing better.
Result
7 of 7
April is doing better because she has higher percentile $(0.99>0.93)$.
Exercise 130
Step 1
1 of 1
Problems involving statistics seem to make the most problems because they are different from everything else in math. They are a separate kind of problems with their own themes and concepts.
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