## SCIE 4101 Review Packet #2 Solutions March 19, 2013

Posted by Ms. Miller in UTA.

Please let me know if you see any errors.

1. We want the relationship between the terms and their positions. We can see that each term goes up by 5, so that is our “slope”, and if we subtract 5 from the first term, that gives us our “y-intercept”: $5n-2$.
2. This is not a function because an element in the domain set is mapped to more than one element in the codomain set.
3. To write the equation, we first must find the slope. Pick two points and plug into the slope formula, for example, (2,5) and (4,13):
$m=\frac{13-5}{4-2}=\frac{8}{2}=4$
Then, we take our slope and use either the point-slope form or the slope-intercept form to write the equation:

Point-Slope Form Slope-Intercept Form
$y-y_{1}=m(x-x_{1})$
$y-5=4(x-2)$
$y=4x-3$
$y=mx+b$
$5=4(2)+b$
$b=-3$
$y=4x-3$
4. To find the set for the independent variable, we set the function equal to each member of the dependent variable and solve for x:
 $x^2-8=-8$ $x^2=0$ $x=0$ $x^2-8=-7$ $x^2=1$ $x=\pm 1$ $x^2-8=8$ $x^2=16$ $x=\pm 4$ $x^2-8=17$ $x^2=25$ $x=\pm 5$

Therefore, the set for the independent variable is $\{0, \pm 1, \pm 4, \pm 5\}$.

5. The cost of electricity decreases $1580 for every 3 degrees rise in temperature. Therefore, if the thermostat is set at 80° (2*3°), the cost would decrease$3160 (2*1580), so a reasonable cost would be $9510-3160 = 6350$.
6. To graph each equation, plot the y-intercept and then use the slope to find a second point:

a.
b.
c.

7. To solve this problem, we need to set up a system of equations. The first equation is the relation between the values of the coins: $0.10d+0.25q=3.20$. The second is the relation between the quantities: $d+q=20$. I would use substitution to work this: rewriting the second equation as $d=20-q$ and substitute it for $d$ in the first equation: $0.10(20-q)+0.25q=3.20$. Then, solve for $q$: $q=8$. Substitute this back into the first equation and solve for $d$: $d=20-8=12$. Thus, Lindsey had 8 quarters and 12 dimes.

1. So the solution is $(-3,2)$.
2. I would use either elimination or matrices:
$3x+2y=21$
$\underline{4x-2y=14}$
$7x=35$
$x=5$
Substitute this into one of the equations to find y:
$3(5)+2y=21$
$15+2y=21$
$2y=6$
$y=3$
Therefore, the solution is $(5,3)$.
3. You can either rewrite the first equation into “y=” form and graph, you can rewrite the second to $3x+y=-5$ and use elimination, or you can substitute the second into the first, solve for x, and then substitute it back to solve for y. In any event, the solution is $(-\frac{6}{5},-\frac{7}{5})$.
8. To write an inequality from a graph, first determine the equation for the line: the slope is 2, and the y-intercept is -3, so the equation of the line is $y=2x-3$. It is shaded above the line, and the line is solid, so we know it is greater than or equal to. Therefore, the inequality is $y\geq 2x-3$.
9. To graph a system of inequalities, graph each inequality, and the solution is the intersection:
10. To graph these inequalities, first solve for y:
$y<-x+6$
$y>\frac{3}{4}x-2$

11. Direct variation is of the form $y=kx$ and inverse variation is of the form $y=\frac{k}{x}$.
1. As x increases, y increases, so this is a direct variation. Substituting into the formula, we have $6=2k$, which means that $k=3$.
2. As x increases, y decreases, so this is an inverse variation. Substituting, we have $12=\frac{k}{0.2}$, which means that $k=2.4$.
12. Substitute this into the inverse variation form: $12=\frac{k}{8}$, which means that $k=96$.
13. If we plot the points, we can see that the coordinates for point A are $(2,6)$:

To write the equation of a line from two points, we first find the slope: $m=\frac{2-6}{5-2}=\frac{-4}{3}=-\frac{4}{3}$. Next, we can use either the slope-intercept form and find the y-intercept or use the point-slope form:
$y-2=-\frac{4}{3}(x-5)$
$y=-\frac{4}{3}x+\frac{20}{3}+\frac{6}{3}$
$y=-\frac{4}{3}x+\frac{26}{3}$

15.

 a. $2x^3+4x^2+4x-7$ b. $s^2-3t^2$

16.

 a. $20m^5$ b. $-12x^4y^7$ c. $-3a^5b^3c^2$

17.

 a. $5r^3s^2-15r^2s^3$ b. $-10x^3+15x^2+5x$ c. $10x^5y^5-5x^4y^6$

18.

 a. $x^2+x-6$ b. $7x^2-52x-32$ c. $m^2-10m+25$ d. $4a^2-9b^2$ e. $9c^2-48cd+64d^2$ f. $12x^2-35x+8$ g. $25g^6-20g^3+4$ h. $9t^4-100$

19.

 a. $5y(2y^2+4y-1)$ b. $-4x(3+2x)$ c. $7ab^2(2ab^2+9b-1)$

20.

 a. $(x+5)(x+2)$ b. $(x-4)(x-2)$ c. $(x+9)(x-2)$ d. $5x^2-10x-4x+8$$5x(x-2)-4(x-2)$$(5x-4)(x-2)$ e. $3x^2+9x+4x+12$$3x(x+3)+4(x+3)$$(3x+4)(x+3)$ f. $2b^2(5b-8)+5(5b-8)$$(2b^2+5)(5b-8)$ g. $(x-5)^2$ h. $(10x-7)(10x+7)$ i. $(x^2-4)(x^2+4)$$(x-2)(x+2)(x^2+4)$
1. The quadratic function crosses the x-axis at 0 and 8. The coordinates of the vertex are $(4,4)$. The axis of symmetry is a vertical line that goes through the vertex, so its equation is $x=4$.
2. From the equation, $a=5, b=-10,$ and $c=3$. Therefore, the x-coordinate of the vertex is $x=\frac{10}{(2)(5)}=1$. (If we wanted to find the y-coordinate, we would just plug the x-coordinate into the original equation.)
3. The larger the absolute value of the number in front of $x^2$, the narrower the graph: (1) $h(x)$, (2) $f(x)$, (3) $g(x)$.

24.

 a. $x=3,-7$ b. $(x+1)^2=0$$x=-1$ c. $x^2+2x-8=0$$(x+4)(x-2)=0$$x=-4,2$ d. $x^2+12x+36=0$$(x+6)^2=0$$x=-6$ e. $2x^2-2x+5x-5=0$$2x(x-1)+5(x-1)=0$$(2x+5)(x-1)=0$$x=-\frac{5}{2},1$ f. $3x^2-13x+4=0$$3x^2-12x-x+4=0$$3x(x-4)-1(x-4)=0$$(3x-1)(x-4)=0$$x=4,\frac{1}{3}$ g. $x=\frac{4\pm\sqrt{(-4)^2-4(1)(1)}}{(2)(1)}$$x=\frac{4\pm\sqrt{16-4}}{2}$$x=2\pm\sqrt{3}$ h. $(2x+1)^2=0$$x=-\frac{1}{2}$ i. $2x^2+x-7=0$$x=\frac{-1\pm\sqrt{1^2-4(2)(-7)}}{(2)(2)}$$x=\frac{-1\pm\sqrt{1+56}}{4}$$x=\frac{-1\pm\sqrt{57}}{4}$

25.

 a. $x^2+8x+16=0$$D=64-4(1)(16)=0$1 real solution$x=\frac{-8}{(2)(1)}=-4$ b. $D=1-4(0.5)(-3)=7$2 real solutions$x=\frac{-1\pm\sqrt{7}}{(2)(0.5)}=-1\pm\sqrt{7}$ c. $-3x^2-2x-1=0$$D=(-2)^2-4(-3)(-1)=-8$No real solutions