Solutions

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1.) $18-(-5)+2\cdot(-6)=18+5-12=11$

2.) $\frac{3}{5}+\frac{1}{6}-\frac{1}{2}=\\\frac{18}{30}+\frac{5}{30}-\frac{15}{30}=\\\frac{8}{30}=\frac{4}{15}$

3.) $(\frac{3}{2}\div\frac{3}{5})-(\frac{5}{6}\cdot\frac{1}{3})=\\(\frac{3}{2}\cdot\frac{5}{3})-(\frac{5}{6}\cdot\frac{1}{3})=\\\frac{15}{6}-\frac{5}{18}=\frac{45}{18}-\frac{5}{18}=\\\frac{40}{18}=\frac{20}{9}$

4.) $\frac{1}{2}\cdot\frac{3}{5}=\frac{3}{10}=30\%$

5.) $3.45+2.74+\frac{11}{5}=3.45+2.74+2.20=8.39\;\text{pounds}$

6.) $40(22.50)+(15\cdot1.5)(22.50)=62.5(22.50)=\1,406.25$

7.) $3,500\cdot10,000=35,000,000=3.5\cdot10^{7}$

8.) Squaring each expression gives the solution: $16.

9.) $\frac{10}{x}=\frac{5}{4}\rightarrow40=5x\rightarrow x=8$

10.) From the text you can derive the proportion: $\frac{x}{27}=\frac{7}{63}\rightarrow x=(27)\frac{7}{63}=3$.

11.) $0.25(0.80)(200)=40$

12.) $\frac{4+5}{12}=\frac{9}{12}=75\%$

13.) Since the mean GPA of the initial 24 student is 3.10, the sum of all their GPA’s must be (24)(3.10). Adding another student with a GPA of 3.60 would make the average:

$\frac{24(3.10)+3.60}{24+1}=3.12$

The question asked for the difference between the old and new average GPA’s which is $3.12-3.10=0.02$.

14.) We must first calculate x:

$\frac{8+9+13+x}{4}=10\rightarrow\\30+x=40\rightarrow\\x=10$

Since the data set has an even number of points, the median is the mean of the middle two points:

$(9+10)/2=9.5$

15.) $((-2)^{2}-1)((-2)-3)=(4-1)(-5)=-15$

16.) First calculate: $MHR=220-36=184$, then substitute MHR and RHR into the equation to solve for THR:

$THR=60+0.75(184-60)=153$

17.) Plug all the values into the equation to get:

$E=(.25)(100)-(.75)(50)=-12.5$

Which means, on average,  you would expect to lose \$12.50 from this bet.

18.) $\frac{2(1200)+x(900)}{2+x}=1140\rightarrow\\2400+900x=2280+1140x\rightarrow\\120=240x\rightarrow\\x=0.5\;\text{hours}$

19.) $(2a+b)(a-3b)=2a^{2}-6ab+ab-3b^{2}=2a^{2}-5ab-3b^{2}$.

20.) $3x^{2}y^{2}+x^{2}y-4xy^{2}-2xy + 2(x^{2}y^{2}-3x^{2}y-xy^{2}+4xy)\rightarrow\\3x^{2}y^{2}+x^{2}y-4xy^{2}-2xy + (2x^{2}y^{2}-6x^{2}y-2xy^{2}+8xy)\rightarrow\\(3+2)x^{2}y^{2}+(1-6)x^{2}y+(-4-2)xy^{2}+(-2+8)xy\rightarrow\\5x^{2}y^{2}-5x^{2}y-6xy^{2}+6xy$.

21.) $x^{2}-14x+49=(x-7)(x-7)=(x-7)^{2}$.

22.) $x-5=-3(1-x)\rightarrow\\ x-5=3x-3\rightarrow\\2x=-2\rightarrow\\x=-1$

23.) $x(x+3)+2=x^{2}+3x+2=(x+1)(x+2)$.

24.) $\frac{5}{3}+\frac{1}{2}=x+\frac{5}{6}\rightarrow\\\frac{10}{6}+\frac{3}{6}=x+\frac{5}{6}\rightarrow\\x=\frac{10+3-5}{6}=\frac{8}{6}=\frac{4}{3}$.

25.) $\frac{(2a)^{3}b^{7}c^{5}}{(2a^{4}bc^{2})^{2}}=\frac{8a^{3}b^{7}c^{5}}{4a^{8}b^{2}c^{4}}=\frac{2b^{5}c}{a^{5}}$

26.) $\sqrt[3]{27x^{6}y^{2}z^{3}}=3x^{2}y^{\frac{2}{3}}z$

27.) $\frac{4\sqrt{x}}{3\sqrt{x}+3\sqrt{y}}\cdot\frac{3\sqrt{x}-3\sqrt{y}}{3\sqrt{x}-3\sqrt{y}}=\frac{12x-12\sqrt{xy}}{9x-9y}=(\frac{4}{3})(\frac{x-\sqrt{xy}}{x-y})$

28.) $\frac{x^{2}-10x+24}{x-4}=\frac{(x-4)(x-6)}{x-4}=x-6$

29.) Perpendicular lines have slopes that are inverses with opposite signs. If we first convert the given equation into slope intercept form, $2x-3y+4=0$ $\rightarrow y=\frac{2}{3}+\frac{4}{3}$, we can see the slope of the perpendicular line will be $-\frac{3}{2}$.

30.) We must first calculate the slope:

$\frac{7-1}{2-0}=\frac{6}{2}=3$

The first point, $(0,1)$, is the y-intercept, so using slope intercept form the equation of the line is:

$y=3x+1$