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Pentagon ABCDE has a vertical line of symmetry. What is the y-coordinate of vertex C so that the area of the pentagon is 40 square units?

Trapezoid ABCD has vertices A(-1,0), B(0,4), C(m,4) and D(k,0), with m>0 and k>0. The line y = -x + 4 is perpendicular to the line containing side CD, and the area of trapezoid ABCD is 34 square units. What is the value of k?

End of the first line, that sequence PE, PO, PF.

Should the O be a D ? Probably, obviously ? it should be.

Should they though be in the order PD, PE, PF, like at the start of the second line ?

If I have to worry about whether or not there are typos in the question, I really can't be bothered with it.

I hope this helps you further.

i guess you can think of it in the form:
ax²+bx+c instead of ax²+bxy+c²

then, factor the resulting quadratic using your methods

yep thats the slope intercept

pretty sure point slope is in the form of
y-y₁=m(x-x₁)
x₁ and y₁ is any point on that line so we can use point (-4,-7) and plug the slope into m to get:

y-(-4)=-6(x-(-7))
y+4=-6(x+7)
is one possible answer


 

Do this:

\(\dfrac{3^{-1}+4^{-1}}{6^{-1}} \hspace{.1cm} =\dfrac {\dfrac {1}{3}+\dfrac {1}{4}}{6^{-1}}\\ \\\:\\ \hspace{.1cm} =\dfrac {\dfrac {7}{12}}{6^{-1}} \hspace{.1cm} =\dfrac {7}{6^{-1}\cdot 12}\\ \\\:\\ \hspace{.1cm}=\dfrac {7}{2}\\\)

GA

where did -4xy & 10xy come from?

3^-1 + 4^-1 means =1/3 + 1/4. But 6^-1 also means 1/6, so:

[1/3+1/4] / 1/6=7/12 x 6/1=42/12 =7/2

Break the expression into groups:

(5x^2-4xy) + (10xy-8y^2).

Factor out x and 2y respectively:

x(5x-4y) + 2y(5x-4y).

Factor out common term:

(x+2y)(5x - 4y) <---- Answer

No, that is not right!. You can only add exponents if they are multiplied together. But you have + signs on top. The right answer is: x^(2 n - 1) + x^(3 n + 4) + x^(4 n + 9). That is it. You can't do much else with it.

A sphere is inscribed inside a hemisphere of radius 2. What is the volume of this sphere?

I think the radius of tthe sphere would be 1.

So 

\(V=\frac{4}{3}\pi*r^3\\ V=\frac{4}{3}\pi*1^3\\ V=\frac{4}{3}\pi\\ V\approx 4.2\; units ^3\)

Ok, thanks so much!!!

This is actually a very easy question.

Multiply -5x+10y=10 by -2 because we want something common with the first equation and negative because we want to eliminate it later.

We get 10x-20y=-20

Then we cross out the 20y

And we are left with: -9x=22 and 10x=-20

Then we add them(both sides) and get x=2

And then we substitute it in whichever equation you want to

-9x+20y=22

-9(2)+20y=22

-18+20y=22

Add 18 on both sides

20y=40

Then divide 20 on both sides

y=2

So we get x=2 and y=2!

Then you can substitute them back into both equations and check your work but I don't have time so bye. Hope that helps!! 

I just did my calculations wrong on one of them, thanks Melody for reminding me!

Both pairs of opposite sides are congruent

Both pairs of opposite sides are parallel

The common ratio is 1.5 for every 5th term. Since there are:

43 - 28 =15 terms. Therefore:

1.5^3 x 43 =81 - 43rd term.

175 miles

>140 miles.

12\( \)

A group of 68 people attend a ball game. There were three times as many children as adults in the group.
Write a system of equations that you could use to set up this problem,
where a is the number of adults and c is the number of children in the group.
Solve the system of equations for c, the number of children in the group.

Let a = adults
Let c = children

\(\begin{array}{|lrcll|} \hline (1) & a + c &=& 68 \\ & a &=& 68-c \\\\ (2) & c &=& 3a \quad & | \quad a = 68-c \\ & c &=& 3(68-c) \\ & c &=& 204-3c \quad & | \quad + 3c \\ & c+3c &=& 204 \\ & 4c &=& 204 \quad & | \quad : 4 \\ & \mathbf{ c} & \mathbf{=}& \mathbf{51} \\\\ & a &=& 68-c \\ & a &=& 68-51 \\ & \mathbf{a} & \mathbf{=}& \mathbf{17} \\ \hline \end{array} \)

The number of children in the group is 51