Help ASAP Please with this math problem. Just look at the picture.

Mathematics

1 answer:

kondaur [170]3 years ago

8 0

${( \sqrt{9)} }^{2}$

$={3}^{2}$

$= 9$

${9}^ \frac{1}{2}$

$= ({3}^{2} ) ^{ \frac{1}{2} }$

$= 3 ^{1}$

$= 3$

You might be interested in

Express the function g in the form f ∘ g. (there is more than one correct answer. the function is of the form y = f(g(x)). use n

bija089 [108]

Answer:

1/x+7

Step-by-step explanation:

5 0

4 years ago

3/5 meters per 9 minutes.<br> What is the unit rate in minutes per meter?

shepuryov [24]

Answer:

.06 meters

Step-by-step explanation:

3/5=.6

.6/9=.06

6 0

3 years ago

I need help with my work

zhuklara [117]

The area of the interior above the polar axis is -0.858 square units

<h3>The area bounded by a polar curve</h3>

The area bounded by a polar curve between θ = θ₁ and θ = θ₂ is given by

$A = \int\limits^{\theta_{2} }_{\theta_{1} } {\frac{1}{2}r^{2} } \, d\theta$

Now, since we have the curve r = 1 - sinθ and we want to find the area of the interior above the polar axis, we integrate from θ = 0 to θ = π, since this is the region above the polar axis.

So, $A = \int\limits^{\theta_{2} }_{\theta_{1} } {\frac{1}{2}r^{2} } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}(1 - sin\theta)^{2} } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}[1 - 2sin\theta + (sin\theta)^{2}] } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}\, d\theta - \int\limits^{\pi}_{0} 2sin\theta \, d\theta+ \int\limits^{\pi}_{0} (sin\theta)^{2} } \, d\theta\\$

$A = \int\limits^{\pi}_{0} {\frac{1}{2}\, d\theta - 2\int\limits^{\pi}_{0} sin\theta \, d\theta+ \int\limits^{\pi}_{0} \frac{(1 - cos2\theta)}{2} } \, d\theta\\= \int\limits^{\pi}_{0} {\frac{1}{2}\, d\theta - 2\int\limits^{\pi}_{0} sin\theta \, d\theta+ \int\limits^{\pi}_{0} \frac{1}{2} } \, d\theta -\int\limits^{\pi}_{0} \frac{cos2\theta}{2} } \, d\theta\\$

$A = \int\limits^{\pi}_{0} \, d\theta - 2\int\limits^{\pi}_{0} sin\theta \, d\theta -\int\limits^{\pi}_{0} \frac{cos2\theta}{2} } \, d\theta\\= [\theta]_{0}^{\pi} - 2[-cos\theta]^{\pi} _{0} - [\frac{sin2\theta}{4}]_{0}^{\pi} \\= [\theta]_{0}^{\pi} + 2[cos\theta]^{\pi} _{0} - [\frac{sin2\theta}{4}]_{0}^{\pi}\\= [\pi - 0] + 2[cos\pi - cos0] - \frac{ [sin2\pi - sin0]}{4}\\= \pi + 2[-1 - 1] - \frac{ [0 - 0]}{4}\\= \pi + 2[-2] - \frac{ [0]}{4}\\= \pi - 4 - 0\\= \pi - 4\\= 3.142 - 4\\= -0.858$