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Thank you so much for your work! I really appreciate it :)

Just to make sure

Is it b^3 , or is it b^2?

OK, travis.....I hope my answers were clear  !!!!

thank you Chris

For the last problem  the area of the unshaded portion of the equilateral triangle   is just the area of  a half circle with a radius of 3  =   (1/2)pi * 3^2  =  (9/2)pi  units^2

[ To see this.....each of the congruent unshaded sections is just the area of a 60° sector of a circle with a radius of 3 ]

So  we have    pi  (3)^2  ( 180/360)  =  (1/2) pi * (3)^2  =  (9/2)pi units^2

And the area of the equilateral triangle  =  (1/2) (6)^2 √3 / 2   =  9√3  units ^2

So....the shaded area is just

[9√3  -  (9/2) pi] units^2 =

9 [ √3 - pi/2 ] units^2

I have a question on this problem:\(Find \frac{x}{y} if \begin{align*} 2\sqrt{x} + \frac1y &= 13 \\ 8\sqrt{x} - \frac 3y &= 3. \end{align*}\)

If you can solve this that would be great. Thanks

i do not under stand this

do you see

i am on the internet

but do not have this

on the internet do you see

so do not hesitate to give me a few pointers please

or just help me out do you see

(read this as poetry)

A regular hexagon with sides of 3" is inscribed in a circle. Find the area of a segment formed by a side of the hexagon and the circle.

The circle will also have a radius of 3

The area of the circle will  be  pi (3)^2  =  9 pi units^2       (1)

The area of the hexagon  =    6 (1/4) (3)^2 * √3  =  27√3/4  units^2      (2)

So....the area  we want   is just 1/6 of  (1)  - (2)  =

[ 9pi - 27√3/4 ] / 6  =

[ 36 pi - 27√3] / 24  =

[ 12pi - 9√3] / 8      units^2  =

(3/8) [ 4pi - 3√3 ] units ^2 

This is my HINT for the last question:

Find the area of the triangle

Find the area of the pizza sections (non-shaded)

The angles of a equilateral triangle are always 60 degrees.

That means one of the pizza sections is 1/6 of an entire circle/

Now I will leave the rest up to you.

Find the area of one segment formed by a square with sides of 6" inscribed in a circle.

I'm not exactly sure what this is asking.....if it is asking for the area between one side of the square and the perimeter of the circle, we can proceed as follows :

To find the radius  of the circle  we have that   

2r^2  = 36   divide both sides by 2

r^2  = 18

r = √18

So....the area we are looking for is the area of a quarter circle with a radius of √18 minus the area of an isosceles right triangle with legs of √18  and a hypotenuse of 6

So....the area is

pi (r^2) /4    -   (1/2) (product of leg lengths)  =

pi (√18)^2 / 4  - (1/2)(√18)(√18)  =

18 ( pi /4  - 1/2)   units^2

how do u do math    

Sometimes I smoke it but usually I just chop it up real fine and...  oh wait, you said math... never mind. 

if i take four letters out of a word. What word do I get? 

Is this a trick question?

Consider it literally.  Write it as follows: 

If I take four letters out of "a word" what word do I get? 

Take out the four letters w-o-r-d and the word you're left with is a. 

.

Yes, but on number 10 I come out with a decimal. Is it suppose to be a decimal?

There are THREE @   16.4   oz       only two at 15.6    and two at 17.7  and none at 21.8 oz

Add all of the measurements....they have to total 360 degrees....solve for 'x'

9x-22 +61+5x-7+34 = 360       does that help?

10 + 2 × (3 ^ 2 - 2)

=    10 + 2 × (9 - 2)

=    10 + 2 × 7

=    10 + 14

=    24

Anytime you forget exponent rules, try writing the whole thing out like this:

=   (2a³)⁴

=   (2a³)(2a³)(2a³)(2a³)

=   (2aaa)(2aaa)(2aaa)(2aaa)

=   (2 * 2 * 2 * 2) * (aaa * aaa * aaa * aaa)

=   2⁴ * a¹²

=   16a¹²

And here's the shortcut method:

=   (2a³)⁴

=   2⁴ a^{(3 * 4)}

=   2⁴ a¹²

=   16a¹²

ordinals are able to be sorted

without getting too smart-arsed I'd say D is the only ordinal variable

plz im stumped sombody answer this

The total length of the rectangle is 10

35-41                                         5

42-48                                       31

49-55                                       19

                                             ______

                                                 55%

So   .55 * 150   =  82.5  ≈   83

(8x^3−6x^2+6x−5) /  (2x^2−1 )

Divide the polynomials....disregard any remainder

                      4x   - 3

2x^2  - 1    [   8x^3 -  6x^2  + 6x  - 5  ]

                      8x^3               -4x

                    __________________________

                               -6x^2           -5

                               -6x^2          + 3

                            __________________

                                              10x  - 8

The slant asymptote  is   y =  4x - 3

This graph confirms this :    https://www.desmos.com/calculator/5xmjxmfwsh

what do you think about that ?