Question

6. Evaluate the function at each specified value of the independent variable and simplify. (If an answer is undefined, enter UNDEFINED.) 7. Evaluate the function at each specified value of the independent variable and simplify. (If an answer is undefined, enter UNDEFINED.)

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

6a

    $V(3) = 113.112$

6b

     $V(\frac{3}{2}) = 14.139$

6c

   $V(2r) = 33.514r^3$

7a

  $h(5) = 0$

7b

    $h(t) = -5.76$

7c

$h(x+5) = x(x+5)$

Step-by-step explanation:

Considering the question 6

The function given is $V(r) = \frac{4}{3} \pi r^3$

For  V(3) we have  

        $V(3) = \frac{4}{3} \pi 3^3$

       $V(3) = \frac{4}{3} * 3.142 * 3^3$

       $V(3) = 113.112$

For  $V(\frac{3}{2})$

        $V(\frac{3}{2}) = \frac{4}{3} * \pi * (\frac{3}{2})^3$

        $V(\frac{3}{2}) = \frac{4}{3} * 3.142 * (\frac{3}{2})^3$

        $V(\frac{3}{2}) = 14.139$

For   V(2r)

            $V(2r) = \frac{4}{3} \pi (2r)^3$

           $V(2r) = \frac{4}{3} * 3.142*8r^3$

           $V(2r) = 33.514r^3$

Considering the question 7

     The function given is  $h(t) = t^2 - 5t$

For  h(5)

          $h(5) = 5^2 - 5(5)$    

           $h(5) = 0$  

For  h(1.8)

            $h(t) = 1.8^2 - 5(1.8)$

            $h(t) = -5.76$

For  h(x+ 5)

              $h(x+5) = (x+5)^2 - 5(x+5)$

               $h(x+5) = x^2 + 25+10x-5x-25$

             $h(x+5) = x(x+5)$