hectictar

The sphere and the wire will have the same volume.

volume of sphere  =  4/3 * pi * 3³  =  4/3 * pi * 27  =  36pi

volume of wire  =  area of cross section * length  =  pi * (0.4/2)² * length  =  0.04pi * length

volume of sphere  =  volume of wire

36pi  =  0.04pi * length                       Divide both sides of this equation by  0.04pi .

900  =  length

So the wire's length is  900 cm .

a)  f(x)  =  \(\frac95\)x - 4          This has an inverse because it is just a linear equation.

y  =  \(\frac95\)x - 4        To find the inverse, solve this equation for  x , so add  4  to both sides.   

y + 4  =  \(\frac95\)x       Multiply both sides by  \(\frac59\) .

\(\frac59\)(y + 4)  =  x     So the inverse function is...

f^-1(x)  =  \(\frac59\)(x + 4)      And to find  f^-1(20) , plug in  20  for  x  into this function.

f^-1(20)  =  \(\frac59\)(20 + 4)   =   \(\frac59\)(24)   =   \(\frac{40}3\)

b)  g(x)  =  4x² + 8x + 13

g(x)  does not have an inverse function because it would have two different  y  values for an  x  value, and for an equaton to qualify as a function, there can only be one  y  value for every  x  value.

c)  h(x)  =  \(\frac1{\sqrt{x}}\)  for  x > 0        Yes this has an inverse.

y  =  \(\frac1{\sqrt{x}}\)             To find the inverse, solve this equation for  x .

y\(\sqrt{x}\)  =  1

\(\sqrt{x}\)  =  \(\frac1{y}\)                 Square both sides.

x  =  \(\frac1{y^2}\)                    So the inverse function is..

f^-1(x)  =  \(\frac1{x^2}\)   for   x > 0

f^-1(4)  =  \(\frac{1}{4^2}\)   =   \(\frac{1}{16}\)

\(\pi^3\)   ≈   31.01

Is this the right expresion?

\(\frac{y^2+2y}{x^2+2xy+y^2}\div\frac{y^2-4}{x+y}\)

First let's factor the numerators and denominators.

\(=\,\frac{y(y+2)}{x^2+xy +xy+y^2}\div \frac{(y+2)(y-2)}{x+y} \\~\\ =\,\frac{y(y+2)}{x(x+y) +y(x+y)}\div \frac{(y+2)(y-2)}{x+y} \\~\\ =\,\frac{y(y+2)}{(x+y)(x+y)}\div \frac{(y+2)(y-2)}{x+y}\)

Now Invert the second fraction and change to multiplication.

\(=\,\frac{y(y+2)}{(x+y)(x+y)}\cdot \frac{x+y}{(y+2)(y-2)} \\~\\ =\,\frac{y(y+2)(x+y)}{(x+y)(x+y)(y+2)(y-2)}\)

Now reduce the fraction as much as possible.

\(=\,\frac{y}{(x+y)(y-2)} \)

y  =  x³ + ax² + bx + 3

We know that when  x = 2  , y''  =  0 .

y'  =  3x² + 2ax + b

y''  =  6x + 2a                So....

0  =  6(2) + 2a

a  =  -6

Since the graph passes through  (2, -1) , we know...

-1  =  2³ + (-6)(2)² + b(2) + 3

-1  =  8 - 24 + 2b + 3

12  =  2b

b  =  6

2/3  =  (4/3) / 2

So he needs 4/3 scoops. That is 1 and 1/3 scoops.

I think the quotient is  x + 2  and the remainder is  6  ,

but I'm not sure on that definition of quotient.

First let's draw two lines perpendicular to the base, forming two right triangles, like this...

Now we can find  h  and  x  .

sin 60°  =  h / 2          so          h   =   2 sin 60°   =   √3

cos 60°  =  x / 2          so          x   =   2 cos 60°   =   1

So...

the height of the trapezoid   =   √3   ,

the first base   =   2  ,   and

the second base   =   2 + 1 + 1  =  4

area of trapezoid  =  (first base + second base)/2  *  height

area of trapezoid  =  (2 + 4)/2  *  √3

area of trapezoid  =  3√3    sq. units

The sum of the interior angles in every quadrilateral  =  360°

So...

48° + 90° + 60° + x°  =  360°

198° + x°  =  360°           Subtract  198°  from both sides.

x°  =  360° - 198°

x°  =  162°