Hello,
$Area= \int\limits^a_b {\rho^2} \, d\theta$

$\frac{1}{2} \int\limits^{\frac{\pi}{5}} _0 {(3\ sin(5\theta))^2} \, d\theta\\\\
=\frac{9}{4} \int\limits^{\frac{\pi}{5}} _0 {(1-cos(10\theta))} \, d\theta\\\\
=\frac{9}{4} [\theta-\frac{sin(10\theta)}{10}]^\frac{\pi}{5}} _0 \\\\
=\frac{9\pi}{20}$

≈1.41371669412

3 0

3 years ago

If g(x) = -2x+7, what is the new function defined by h(x) = g(2x^2 +1)

hammer [34]

Answer:

h(x) = - 4x² + 5

Step-by-step explanation:

To find h(x) = g(2x² + 1), substitute x = 2x² + 1 into g(x)

h(x) = - 2(2x² + 1) + 7

= - 4x² - 2 + 7

= - 4x² + 5

5 0

4 years ago

20 104030.1540702520301004050​

20 104030.1540702520301004050