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You know Chris,

in Desmos where it says "share" one of the options is that you can share as an image.

You can paste the URL that is thrown up into the image thing here and the pic comes straight up :)

Mountains Out Of Mole Hills.........indeed.......LOL!!!!!!!

Thanks, Melody......

what is the answer to this math problum? i am confused: a+b2*c6 / =6 c2

Well my first job is to interpete what your question is :)

\( a+b^2*c^6 =6 c^2\\\)

NOPE  I think you are just playing :))

In how many ways can the letters of the word CLASSROOM be arranged such that the new word does not begin or end with a vowel?

Hi Chris   

How about I do it a different way and we will see if our answers are the same  

The number of arrangements in total, if there were no restrictions is

\(\frac{9!}{2!2!}\)

9!/(2!*2!) = 90720

​

How many ways can it begin with a vowel

If it starts with an O then 8!/2! = 20160 ways

If it starts with an A then 8!/(2!2!) = 10080ways

It is the same for ending in a vowel so that is

2*(20160+10080) = 60480

BUT

there is double counting happening here because is it begins and ends in a vowel it has been inclued both times.

SO how many start AND end in a vowel;?

Start with A end in O =  7!/2! = 2520 ways

Start with O end in A =   2520 ways

Start with O end in O = 2520 ways

That is    3*2520 = 7560 ways

So altogether it seems that   60480-7560 = 52920  either start or end with a vowel.

So the number that DO NOT start OR end in a vowel is    90720-52920 = 37800

YEP we can't both be wrong....   LOL 

See I have been studying very hard!!

-20x+-12

-4*5x+3*-4

I think we have

[(√8) x ] / [(³√16 ) x² ] =

[ √8 / ³√16 ] *  [   x  / x²  ]  =

 [ 2√2 / 2 ³√2 ] *  [  1/x ] =

[√2 /  ³√2 ]  * [ 1 / x ]  =

2^(1/2  − 1/3) / x  =

2^ (1/6)  / x   =

⁶√2 / x

I'll give his one a try

We have 6  ways to choose a consonant for the first position and 5 remaining ways to choose a consonant for the last position

And the other letters can be arranged in  7! ways

However........we must divide this by  [2! *2!]   to count only  the "identifiable" words  since we have repeated "O's"  and "S's"

So we have

[ 6 * 5  * 7! ] / [2! * 2!]   = 37800  arrangements

[ Can some other mathematician (Melody??)  check this  ??? ]

The form is   :

4p(x – h) = (y – k)^2

Where  p = 3  , h  =  – 5   and k  =  4    

And we have

4(3)(x + 5)  =  (y – 4)^2

12 (x + 5)  = (y – 4)^2

x + 5  =   (1/12) (y – 4)^2

x   = (1/12) (y – 4)^2  – 5

Here's a graph  :  https://www.desmos.com/calculator/7kcjwmhnxh

\((\frac{11}{5}, \frac{33}{5})\)

I'm assuming no one's helped you yet because the question is too confusing.

A) Is it \(\sqrt[3]{8x} \cdot \sqrt{16x^2}\)

B) Or is it \(\sqrt{8x} \cdot \sqrt{16x^2}\)

C) Or is it \(\sqrt[3]{8x} \cdot \sqrt[3]{16x^2}\)

D) Or is it ​\(\sqrt{8x} \cdot \sqrt[3]{16x^2}\)

E) Or is it neither?

A = \(\sqrt[3]{8} \cdot \sqrt[3]{x} \cdot \sqrt{16} \cdot \sqrt{x} = 2\sqrt[3]{x} \cdot 4x = \boxed{8x\sqrt[3]{x}}\)

B = \(\sqrt{4} \cdot \sqrt{2x} \cdot \sqrt{16} \cdot \sqrt{x^2} = \boxed{8x\sqrt{2x}}\)

C = ​​\(\sqrt[3]{8} \cdot \sqrt[3]{x} \cdot \sqrt[3]{8} \cdot \sqrt[3]{2x^2} = \boxed{4x\sqrt[3]{3}}\)

D = \(\boxed{4\sqrt{2x} \cdot \sqrt[3]{2x^2}}\) 

E = Well, I can't help you there, but next time make your question clearer

You have to imagine the + as +1. So it would be:

\(1(-4) = 1 \times (-4) = -4\)

Because of the Identity Property of Multiplication, anything multiplied by 1 is itself, so it follows for -4 as well.

That means our answer is: \(\boxed{-4}\)

It follows that for any right triangle, you can find the length of the hypotenuse using Pythagorean Theorem. Basically, you add the squares of the legs to get the square of the hypotenuse. If one leg was A, the other was B, and the hypotenuse was C, ​\(A^2 + B^2 = C^2\)

So if A was 3, B was 4, then C would be the square root of 9 + 16 = 25, so the hypotenuse would be 5.

Here's a formula for C, the hypotenuse.

\(C = \sqrt{A^2 + B^2}\)

Find more about the Pythagorean Theorem here:

Here you go:

Work through this and you will find out   :))

anyone PLZ HELP ME

The answer is  5498.12

Mathematics(math for short) is a everyday usage of numbers, such as "I have 25 dollars, and grandma gave me 10, So I have... 35 now!". This is mathematics. There is much, much more to it though. Such as algebraic expressions and harder numeric expressions.

+||I am the one and only REAL MathPeacock||+

If Sam brought 2/3 liters of punch and 3/4 liters of tea, find the least common denominator. Multiply 3 and 4 to find it: 3 * 4 = 12. So now multiply 3 * 3 and 4 * 2. 

This will result in 8/12(2/3) and 9/12(3/4).  Now add:

8/12 + 9/12 = 17/12 = (1 and 5/12 liters)

Sam brought 1-5/12 Liters to the Party.

+||I am the one and only MathPeacock||+

Did you post it? I said I figured it out man. Chill.

What do the last TWO LINES say??

I am gonna call you

Guesty. Like I did before

Oop, Nevermind, I got it now. You are correct.

Lets see. We will use 3.14 as Pi' shortened common.

3.14 * 5 = 15.70

15.70 - 7 = 8.70

8.70 = 8.7 or 'Eight and Seven-tenths'

8.07