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3 years ago

Hi I need help I need to get all a’s and I really need some help

Mathematics

2 answers:

Zepler [3.9K]3 years ago

5 0

Answer:

What exactly do you need help with?

Step-by-step explanation:

Mama L [17]3 years ago

4 0

Answer:

Hi:) Do you have a specific question on something or just some tips for those desired good grades?

Step-by-step explanation:

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A ball has a radius of 18 cm.

jok3333 [9.3K]

Hello there.

A ball has a radius of 18 cm.

What is the approximate volume of the ball?

Use 3.14 to approximate pi. Round to the nearest hundredth if necessary.

Answer: 24416.64

4 0

2 years ago

Read 2 more answers

Find the slope from the pair of points. (0, -5) (-3,5)

bagirrra123 [75]

Answer:

Step-by-step explanation:

your welcome

5 0

3 years ago

HELP ME PLZZZZ!!!!! It’s just 21 and 22

galina1969 [7]

Answer:

21) 12 cm

22) 5.9 cm

Step-by-step explanation:

21) The base is 10, and if the other two sides are congruent and the perimeter is 36, we can figure out with simple algebra that the sides are 13 cm long.

Half of 10 is 5, so we can use the pythagorean theorem.

5^2+x^2=13^2

Rearranging the variables we have 169-25=x^2

144=x^2

x can be plus or minus 12, but since negative length is impossible we find that x is positive 12 cm.

22) We want to use sine, because we have opposite and hypotenuse. A simple and easy way to memorize this is the SohCahToa method. If we have opposite (O) and hypotenuse (H) we have OH. Soh has the letters O and H, and the S means we should use sine.

sine 36=a/10

Plug this into a calculator or desmos scientific calculator to get a=5.9

7 0

2 years ago

Rewrite the following integral in spherical coordinates.

lora16 [44]

In cylindrical coordinates, we have $r^2=x^2+y^2$ , so that

$z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}$

correspond to the upper and lower halves of a sphere with radius $\sqrt2$ . In spherical coordinates, this sphere is $\rho=\sqrt2$ .

$1 \le r \le \sqrt2$ means our region is between two cylinders with radius 1 and $\sqrt2$ . In spherical coordinates, the inner cylinder has equation

$x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)$

This cylinder meets the sphere when

$x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)$

which occurs at

$\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi$

where $n\in\Bbb Z$ . Then $\frac\pi4\le\phi\le\frac{3\pi}4$ .

The volume element transforms to

$dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi$

Putting everything together, we have

$\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3$