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The pi thing is not included in the special characters.  How weird is that !!

What is 2/3 x 15

\(15\cdot \frac{2}{3} \\ = \frac{3\cdot 5\cdot 2 }{3} \\ = 5\cdot 2 \\ = 10\)

1/3 of 18

\(18\cdot \frac{1}{3} \\ = \frac{3\cdot 6 }{3} \\ = 6\)

P = 4s    divide both sides by 4

P/ 4   = s

A = P + 1    subtract 1 from both sides

A- 1   = P

C = 2pir    divide both sides by 2pi

C/(2pi)  = r

y = mx + b     subtract mx from both sides

y - mx  = b 

How do I find the perimeter C of this fountain?..

A fountain is made up of two semicircles and a quarter circle.

The diameter of the semicircles is d = 20ft. Find the perimeter of the fountain. Round your answer to the nearest tenth.

\(\large C=\frac{2d\pi}{4}+2\times\frac{d\pi}{2}=\frac{d\pi}{2}+d\pi\) 

Quarter circle + two times semicircle

\(\large C=20ft\times\frac{\pi}{2}+20ft\times \pi=20ft\times\frac{3\pi}{2}\)

\(\large C=94.248ft\)

  !

You can just type that exact thing into the calculator on this website and it will tell you.

It should be \(\frac{10}{3}\)

Unless you want it as a mixed number: \(3\frac{1}{3}\)

If you want to use it in an equation, 10% in equation form is 0.10

... because 10% is out of 100% making that .10

you sphere in a perfect sphere

Either like the Guest said, 2.75. or you can do 2(3/4).

Examples:

a) 2(3/4) x 1(3/5)

b) 2.75 x 1.50

Your welcome, and enjoy your day, evening, or night.

34.9

In trig. how do you find the angle if the opposite side and the adjacent side are given?

Also can you do it in degrees?

This is not solvable.
There must be at least three sizes.

  !

V = (4/3)pi * (.5)^3   ≈  .52cm^3

If you had a bag of cookies and I ate one, and there were only 3 cookies left, how many cookies were in the bag before I ate one?

If you had a bag of cookies and I ate 28 of them, and there were only 33 cookies left, how many cookies were in the bag before I ate 28 of them?

Maybe if you think about it in terms of cookies you can figure it out. :)

Just add 28 to both sides... 

Well-presented, hectictar....!!!

(0.0167 cm^-1)^-1

\(\large(0.0167cm^{-1})^{-1}=(\frac{0.0167}{cm})^{-1}=\frac{cm}{0.0167}\)

  !

(25/5)(30) = 150

I remember doing these D:

The slope of AB must match the slope of CD.

The slope of BC must match the slope of AD.

Slope of AB =  \(\frac{-1-6}{-5+9}=-\frac{7}{4}\)

Slope of CD = \(\frac{5+2}{-1-3}=-\frac{7}{4}\)

Slope of BC = \(\frac{6-5}{-9+1}=-\frac{1}{8}\)

Slope of AD = \(\frac{-1+2}{-5-3}=-\frac{1}{8}\)

So far so good. We have just shown that this is a parallelogram at least.

In order for it to be a rhombus:

The slope of BD must be the negative reciprocal of the slope of AC.

Slope of BD = \(\frac{6+2}{-9-3}=-\frac{8}{12}=-\frac{2}{3}\)

Slope of AC = \(\frac{-1-5}{-5+1}=\frac{-6}{-4}=\frac{3}{2}\)

Everything checks out. This figure is infact a rhombus. :)

\(-14=\frac{2(h-3)}{5}\)

Multiply both sides of the equation by 5:

\(-14(5)=2(h-3) \\~\\ -70=2(h-3)\)

Divide both sides of the equation by 2:

\(\frac{-70}{2}=(h-3) \\~\\ -35=h-3\)

Add 3 to both side of the equation:

\(-35+3=h \\~\\ -32=h\)

There are 25 hockey players. There are 5 hockey players per group.

So there are 25/5 = 5 groups.

Each group gets 30 candy bars.

If all 5 groups sold all 30 candy bars, that would be a total of 5*30 = 150 candy bars.

Since it is an isocolese trapezoid, the two nonparallel sides, aka the legs, have the same length.

AB = CD

7y - 4 = 8y - 18

7y +14 = 8y

14 = y

[  I think they just put the info about BC being 4y - 6 to try to trick you. :)  ]

Sure !   Just add the two equations together to get:

8x =16     x = 2

4(2) -9y = 9

8--9y = 9

-9y=1

y = -1/9