Solutions

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1.) $8-15/3+1=8-5+1=4$

2.) $105-(-9)=105+9=114^{\circ}\text{F}$

3.) $\frac{1}{2}+\frac{2}{3}-\frac{3}{4}=\frac{6}{12}+\frac{8}{12}-\frac{9}{12}=\frac{5}{12}$

4.) After the first day $\frac{2}{3}\;$ of the pie is leftover. On the second day $\frac{3}{4}\;$ of this is eaten, leaving $\frac{1}{4}\;$ of $\frac{2}{3}\;$ of the original pie, or: $(\frac{2}{3})(\frac{1}{4})=\frac{1}{6}$.

5.) $\frac{4}{5}\div\frac{2}{3}-\frac{7}{10}=\\\frac{4}{5}\cdot\frac{3}{2}-\frac{7}{10}=\\\frac{12}{10}-\frac{7}{10}=\\\frac{5}{10}=\frac{1}{2}$

6.) $4.6\cdot4.75+82.63=\104.48$

7.) $1.18\;\text{billion}=1,180,000,000=1.18\cdot10^{9}$

8.) $\sqrt{3^{4}}=\sqrt{(3^{2})^{2}}=3^{2}=9$

9.) $\frac{x}{9}=\frac{5}{15}\\\rightarrow \frac{x}{9}=\frac{1}{3}\\\rightarrow 3x=9\\\rightarrow x=3$

10.) $\frac{x}{5}=\frac{6.75}{9}\\\rightarrow 9x=5(6.75)\\\rightarrow x=\3.75$

11.) $0.54(22,000)=11,880\;\text{votes}$

12.) If 60% are over 25, 40% are under 25. Since half of this 40% are under 21, $(0.5)(0.4)=0.2=20\%\;$ are between 21 and 25.

13.) The average age in the room will be the sum of the ages divided by four. Since the sum of ages divided by 10 is equal 14.8, the sum of the ages must be $10\cdot14.8=148$, and therefore the average is $148/4=37$.

14.) $\frac{15(66)+25(62)}{40}=63.5$

15.) $\frac{(4)^{3}+1}{(4)+1}=\frac{65}{5}=13$

16.) $\frac{((5)-(1))^{2}-(-2)}{(-2)(5)-2(1)}=\frac{4^{2}+2}{-10-2}=\frac{18}{-12}=-\frac{3}{2}$

17.) $s=1.15(500-w)$

18.) $\frac{1^{2}+2^{2}+3^{2}+4^{2}+5x}{1+2+3+4+x}=3.75\rightarrow\\30+5x=3.75(10+x)\rightarrow\\30+5x=37.5+3.75x\rightarrow\\1.25x=7.5\rightarrow\\x=6$

19.) $4a^{2}b^{2}-3a^{2}b+ab\;+\;(ab+a^{2}b+5ab^{2}-3a^{2}b^{2})=\\(4-3)a^{2}b^{2}+(-3+1)a^{2}b+5ab^{2}+(1+1)ab=\\a^{2}b^{2}-2a^{2}b+5ab^{2}+2ab$

20.) $5x+1=2(x-4)\\\rightarrow 5x+1=2x-8\\\rightarrow 3x=-9\\\rightarrow x=-3$

21.) $x^{2}-6x+8=(x-4)(x-2)$.

22.) Begin by multiplying both sides of the equation by 3 to get:

$x-\frac{3}{2}=\frac{3}{4}\rightarrow\\x=\frac{3}{4}+\frac{3}{2}\rightarrow\\x=\frac{9}{4}$

23.) $\frac{-2x^{3}y^{2}z}{6x(yz)^{6}}= -\frac{1}{3}x^{3-1}y^{2-6}z^{1-6}=-\frac{1}{3}x^{2}y^{-4}z^{-5}=\frac{-x^{2}}{3y^{4}z^{5}}$

24.) $\frac{x^{2}+6x+5}{x+1}=\frac{(x+1)(x+5)}{x+1}=x+5$.

25.) $\frac{-\sqrt{x}}{2\sqrt{x}-\sqrt{y}}\cdot\frac{2\sqrt{x}+\sqrt{y}}{2\sqrt{x}+\sqrt{y}}=\frac{-(2x+\sqrt{xy})}{4x-y}=\frac{2x+\sqrt{xy}}{y-4x}$.

26.) $\frac{16-x^{2}}{x^{2}+9x+20}=\frac{-(x+4)(x-4)}{(x+4)(x+5)}=\frac{4-x}{x+5}$.

27.) By converting the equation into slope intercept form: $y=-\frac{3}{2}x-3\;$, we can see that the y-intercept occurs at $(0,-3)$.

28.) Let $c\;$ and $k\;$ represent the number of years Carlos and Katie have worked at the company. From the problem we can derive the equations: $c=k+5\;$ and $c=2k$. From these we get:

$k+5=2k\rightarrow\\k=5$

Now substituting this value into either of our original equations yields $c=10$. Hence Carlos has worked at the company for 10 years, and Katie has worked at the company for 5 years.

29.) Solving the system of equations gives the point $(x,y)\;$ of intersection. We can begin by substituting the first equation into the second to get:

$x+(2x-1)=5\rightarrow\\3x=6\rightarrow\\x=2$

Now substituting this value into $y=2x-1\;$ gives:

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

Therefore, the two lines intersect at the point $(2,3)$.

30.) Since the equation is parallel to $y=\frac{1}{2}x$, must have the same slope, so the equation will have the form $y=\frac{1}{2}x+b$. We can solve for $b\;$ by substituting the values of the point $(2,2)\;$ into this equation:

$(2)=\frac{1}{2}(2)+b\\\rightarrow b=2-1=1$

So the equation we’re looking for is $y=\frac{1}{2}x+1$.