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1 quart is 4 cups

so

4 quarts is 16 cups

it ok everyone make mistake and learn from them

There is an error in my solution because I copied your question incorrectly.  

You should be able to fix it Shaniab.  I am on my phone and it is all too difficult from here :(

I had x instead of 4x  :(

i know right

Maybe but you can learn here    

i know i am sorry ok just trying

i think my  teacher i teching wrong

Thanks Chris  

By the way these are airports for model planes.  Either that or it is Lilliput    

No sasini      

thje first line should be

-c=-13-7

I cannot make sense of what you have done Shaniab.  

Remember - one equal sign per line - NEVER MORE  

So 

$$x=\sqrt[3]{9}$$

Your answer appears to bre perfect Sasini.  Thank you  

Shaniab, It is important that when you show woking for your answers that you only ever put one equal sign per line.

It is too hard to think it through otherwise :)

1 -5(x^3+x^2) = -5x^2 -4(x^3 +2)

1 -5x^3 -5x^2  = -5x^2 -4x^3 -8

1 -5x^3            =  -4x^3 - 8

1 +8                 =   -4x^3  + 5x^3

9                      =      x^3

1−5(x3+x2)=-5x2−4(x3+2)1−5x3−5x2=-5x2−4x3−81−5x2=-5x2+x3−89−5x2=-5x2+x39=x327=x

-48+9(x4−4x)=6(12−x)+9x4-48+9x4−36x=72−6x+9x4-48+9x4=72+30x+9x4-120+9x4=30x+9x4-120=30x-4=x

7 - c = -13

-c     = -13 + 7

-c     =  -6

-c/-1 = -6/-1

 c      = 6//

When you write these Shaniab you should use the 'hat' to indicate exponents. :)

1−5(x^3+x^2)=-5x^2−4(x^3+2)

I would like you to try and do it yourself Shaniab

Just remember that the minus sign belongs to the number that follows it.

I have added stuff  that might help you a bit.

1+[−5(x^3+x^2)]=-5x^2+[−4(x^3+2)]

Now can you give it a go?

You are welcome :)

It doesn't equal zero

$$10^{-156}=\dfrac{1}{10^{156}}$$

This number is so tiny that most calculators cannot express it as a number

10^156 is 1 followed 156 zeros

That is a hubge number.

1 divided by a huge number is miniscule    

thank u

$$\\-48+9(x^4-x)=6(12-x)+9x^4\\
-48+9x^4-9x=72-6x+9x^4\\
-48-9x=72-6x\\
$add 6x to both sides$\\
-48-3x=72\\
$add 48 to both sides$\\
-3x=120\\
x=-40$$

if it is in the form     $$(x-h)^2=4a(y-k)$$

then the vertex is (h,k)

The focal length is    $$|a|$$              (note that a can be negative)

So the focus will be     $$(h, k+a)$$

it can be negative too and it can be 0.

If you have the equation of a parabola in vertex form y = a(x - h)² + k, then the vertex is at (h, k) and the focus is (h, k + 1/(4a)). Notice that here we are working with a parabola with a vertical axis of symmetry, so the x-coordinate of the focus is the same as the x-coordinate of the vertex

Excellent SevenUp       

I here are the original  answers