Question

Help me with these math questions.......

Answer 1

Answer: 118.40, 300, $18,000

Step-by-step explanation:

a) p(x) = $-\frac{1}{5} (8) + 120$

   p(8) =  $-\frac{8}{5} + 120$

          = -1.60 + 120

          = 118.40

b) R(x) = x * p(x)

         = $x(-\frac{1}{5}x + 120)$

         = $-\frac{1}{5}x^{2} + 120x$

           a=$-\frac{1}{5}$, b=120

x = $\frac{-b}{2a}$

  = $\frac{-120}{-\frac{2}{5}}$

  = $\frac{-120(5)}{-2}$

  = 300

c)  R(x) = $-\frac{1}{5}x^{2} + 120x$

R(300) = $-\frac{1}{5}(300)^{2} + 120(300)$

           = -18,000 + 36,000

           = 18,000

***********************************************************************

sec θ = $\sqrt{6}$

$\frac{hypotenuse}{adjacent} = \frac{\sqrt{6}}{1}$

adjacent² + opposite² = hypotenuse²

      1²       + opposite² =      (√6)²

      1         + opposite² =        6

                   opposite² =        5

                   opposite =        √5

csc θ = $\frac{hypotenuse}{opposite} = \frac{\sqrt{6}}{\sqrt{5}} = \frac{\sqrt{30}}{5}$

cot θ = $\frac{adjacent}{opposite} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$

sin θ = $\frac{opposite}{hypotenuse} = \frac{\sqrt{5}}{\sqrt{6}} = \frac{\sqrt{30}}{6}$

cos θ = $\frac{adjacent}{hypotenuse} = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6}$

tan θ = $\frac{opposite}{adjacent} = \frac{\sqrt{5}}{1} = \sqrt{5}$

**********************************************************************

Answer: 144°

Step-by-step explanation:

$\frac{\pi}{180}=\frac{4\pi}{5(x)}$

π(5x) = 180(4π)

    x   = $\frac{180(4\pi)}{5\pi}$

         = 36(4)

         = 144