gwenspooner85

a+b=2

a+c=7

b+c=9

Add all three equations together to get 2a+2b+2c=18. We divide this equation by 2 to get a+b+c=9.

1.

 $2x-5\le-x+12\\ =3x\le17\\ =x\ge\frac{17}{3}\\ \boxed{x\in(-\infty, \frac{17}{3}]}$

2. $-4(x+4)>x+7\\ =-4x-16>x+7\\ =5x<-23\\ x<-\frac{23}{5}\\ \boxed{x\in(-\infty, -\frac{23}{5})}$

It's just two, not a big deal in my opinion. Any more than that would be an overload.

From numbers 1-9, 9 digits are used. From numbers 10-99, 90*2=180 digits are used. From numbers 100-999, 900*3=2700 digits are used. The number of pages in this book must fall somewhere between 100-999. We already have 180+9=189 digits. 

1629-189=1440.

We divide 1440 by 3 to find the number of pages between 100-999. We get 480 pages.

We use the equation a+d(n-1), where a is the first term and d is the common difference, and n is the nth term. 

We form two equations: a+d(1)=-2 and a+d(4)=16. We have a system of equations and solve.

a+d=-2

a+4d=16

-3d=-18

d=6, a=-8

So the fourteenth term is -8+6(14-1)=-8+6(13)=-8+78=70.

A: 2/5

B: 4/5

So I think B is more likely to happen.

My counting and probability is not very good but I'll try. I think this is how to do the problem:

We look at where the teacher chooses all girls in the group. It's a group of 3 and there are 4 girls so there are 4C3=4 ways to choose 3 girls(a group). Then we look at the total number of groups the teacher can choose which is 10C3=120. Therefore the probability everyone in the group is a girl is 4/120=1/30.

We simplify all of these expressions.

A:

$A=\frac{2^{\frac{1}{2}}}{4^{\frac{1}{6}}}\\ A=\frac{2^{\frac{1}{2}}}{(2^2)^{\frac{1}{6}}}\\ A=\frac{2^{\frac{1}{2}}}{2^{\frac{1}{4}}}\\ A=2^\frac{1}{4}$

B:

$B=\sqrt[12]{128}\\ B=\sqrt[12]{2^7}\\ B=2^\frac{7}{12}$

C:

$C=(\frac{1}{8^\frac{1}{5}})^2\\ C=(\frac{1}{(2^3)^\frac{1}{5}})^2\\ C=(\frac{1}{2^\frac{3}{5}})^2\\ C=(2^{-\frac{3}{5}})^2\\ C=(2^{-{\frac{6}{5}}})\\ $

D:

$D=\sqrt{\frac{4^{-1}}{2^{-1}\cdot8^{-1}}}\\ D=\sqrt{\frac{(2^2)^{-1}}{2^{-1}\cdot(2^3)^{-1}}}\\ D=\sqrt{\frac{2^{-2}}{2^{-4}}}\\ D=2$

E:

$E=\sqrt[3]{2^\frac{1}{2}\cdot4^-{\frac{1}{4}}}\\ E=\sqrt[3]{2^\frac{1}{2}\cdot(2^2)^-{\frac{1}{4}}}\\ E=\sqrt[3]{2^\frac{1}{2}\cdot2^-{\frac{1}{2}}}\\ E=\sqrt[3]{1}\\ E=1$

We then order them: C, E, A, B, D.

We complete the square to get $y=-2(y-3)^2+3$. The vertex is (h, k). Therfore the vertex of the parabola $x = -2y^2 + 12y - 15\ \text {is} \ \boxed{(3, 3)}.$

The line perpendicular to y=2x+5 must have the slope -1/2. We then have that the perpendicular line's equation is y=-1/2x+b. We plug in the given point (5, 5) to solve for b.

y=-1/2x+b

5=-1/2(5)+b

5=-5/2+b

b=15/2

So the equation of this line is y=-1/2x+15/2. To find the intersection point of it and y=2x+5, we set -1/2x+15/2 and 2x+5 to be equal to each other.

-1/2x+15/2=2x+5

5/2x=5/2

x=1

Plug this value into either of the equations for the lines we have. y=2x+5 will be easier to work with.

y=2x+5

y=2(1)+5

y=7

Therefore the intersection point is (1, 7).