hectictar

I want to know how to do this problem too.....

Thanks Melody!! Now, we need a new question!! 

This is a funny lookin pyramid! The formula for the volume of this pyramid is the same as if "P" were above the center of the base:  volume = (1/3)(area of base)(height)

surface area  =  4 * π * (radius)²

And they tell us that the

radius  =  2/3 t³

And we want an equation that gives the surface area for any value of  t  ( that is ≥ 0 ) .

So....instead of calling the radius "radius", we want to say the radius is  2/3 t³  .

surface area  =  4 * π * (2/3 t³)²          Then simplify this equation.

surface area  =  4 * π * 4/9 * t⁶

surface area  =  16 π t⁶ / 9             So we can say...

s(t)  =  16 π t⁶ / 9

Here's one way....

f(x)  =  ax⁴ - bx² + x + 5                        Plug in  -x  for  x  .

f(-x)  =  a(-x)⁴ - b(-x)² + (-x) + 5             (-x)⁴  =  x⁴     and     (-x)²  =  x²

f(-x)  =  ax⁴ - bx² - x + 5                        Rearrange.

f(-x)  =  ax⁴ - bx² + 5 - x                        Add  x  and subtract  x  .

f(-x)  =  ax⁴ - bx² + x + 5 - x - x            Substitute  f(x)  in for  ax⁴ - bx² + x + 5  .

f(-x)  =  f(x) - x - x          Plug in  3  for  x .

f(-3)  =  f(3) - 3 - 3         Substitute  2  in for  f(-3)  and solve for  f(3) .

2  =  f(3) - 3 - 3

2  =  f(3) - 6

8  =  f(3)

\(\sqrt{432}\,=\,\sqrt{2*2*2*2*3*3*3}\,=\,\sqrt{2*2}*\sqrt{2*2}*\sqrt{3*3}*\sqrt3\,=\,2*2*3*\sqrt3\)

\(=\,12\sqrt3\)          Right!

Is this the third one??

\(\sqrt[3]{125x^{21}y^{33}}\,\,=\,(125x^{21}y^{33})^{\frac13}\,=\,125^{\frac13}x^{\frac{21}{3}}y^{\frac{33}{3}}\,=\,5x^{7}y^{11}\)

\(9\sqrt{56x^7y^{12}}\,=\,9\sqrt{2*2*2*7*x*x^6*y^{12}}\,=\,9*\sqrt{2*2}*\sqrt{2*7*x}*\sqrt{x^6}*\sqrt{y^{12}}\\~\\=\,9*2*\sqrt{14x}*x^3*y^6\,=\,18x^3y^6\sqrt{14x}\)

the length of PR  =  the length of PN + the length of NR

the length of PR  =  4x + 5   +   7x - 16

the length of PR  =  4x + 7x + 5 - 16 

the length of PR  =  11x  - 11

\(\frac{x-1}{x-2}-\frac{2}{x}\,=\,\frac{1}{x-2}\)

First let's get a common denominator.

\(\frac{(x)(x-1)}{(x)(x-2)}-\frac{2(x-2)}{(x)(x-2)}\,=\,\frac{1x}{(x)(x-2)}\)

We can multiply through by the denominator....however....you are right that x can't be 2 !

Also  x  can't be  0 . If we plug in  0  or  2  for  x  into the original equation, we will get an undefined result. Before we multiply through by the denominator, we must say that  x ≠ 0  and  x ≠ 2 .

\((x)(x-1)-2(x-2) \,=\,1x\)         Distribute the  x  and the  -2 .

\(x^2-x-2x+4 \,=\,x\)           Subtract  x  from both sides and combine like terms.

\( x^2-4x+4 \,=\,0\)          Factor the left side and solve for  x  .

\((x-2)^2=0 \\~\\ x-2=0 \\~\\ x=2\)

But, x cannot be 2 because it causes a zero in the denominator of the original problem.

So there is no solution.

Hey!! 

Let's get this equation of a circle into the form  (x - h)² + (y - k)²  =  r²   , where  (h, k)  is the center of the circle, and  r  is its radius.

x² + y² - 7  =  4y - 14x + 3

                                                  Add  7  to both sides.

x²  +  y²  =  4y - 14x + 10

                                                 Add  14x  to both sides, and subtract  4y  from both sides.

x² + 14x   +  y² - 4y    =  10

                                                 Add  49   and add  4  to both sides to complete the squares.

x² + 14x + 49  +  y² - 4y + 4   =   10 + 49 + 4

                                                                         Factor  x² + 14x + 49  as a perfect square trinomial.

(x + 7)²   +   y² - 4y + 4  =  63

                                                                         Factor  y² - 4y + 4  as a perfect square trinomial.

(x + 7)²   +   (y - 2)²  =  63

Now that the equation is in this form, we know that..

r²  =  63

And...

area of circle    =    pi * r²    =    pi * 63    =    63pi