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alexandr1967 [171]
2 years ago
10

8.

Mathematics
1 answer:
svp [43]2 years ago
7 0
A 80 square meter I’m not quite sure
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Please help! Create a set of five positive numbers (repeats allowed) that have a median of 10 and a mean of 14. What thought pro
Brilliant_brown [7]
Alright, since there are 5 numbers, and the mean (or average) is (sum)/(amount of numbers), we have (sum)/5=14. Multiplying both sides by 5, we have the sum being 80. The median of 10 means that in a, b, c, d, e, 10 has to be c and the numbers have to be in ascending order. A and b must be 10 or lower, while d and e must be 10 or higher. Putting some random numbers in, we can have 1, 1, 10, 15, and e. We left e there because the sum needs to be 80, and since 1+1+10+15=27, 80-27=53=e. This, however, would not work if e was less than 10 and we therefore would have needed to make some numbers lower to compensate for this. Our answer is therefore 1, 1, 10, 15, 53
6 0
3 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

4 0
2 years ago
Solve the following quadratic equation for its exact value by using the square root principle.
Korvikt [17]
Idk what the hell all that is but x = 1.802775637731995

4x^2 - 13 = 0

4x^2 = 13

13/4=3.25

x^2 = 3.25

x = ~1.8
5 0
3 years ago
PLEASE HELP ME!!!!
antiseptic1488 [7]

Answer:

OBC= 15

OCR =30

In OBQ

OBQ + QOB + OQB = 180

OBQ= OBC=15

OQB= 90

SO, QOB = 180- OBQ- OQB

= 180-15 -90

= 75

c) is correct

5 0
3 years ago
Read 2 more answers
The speedy punt returner and the ferocious defensive player are headed straight toward each other. The punt returner is travelin
Svetach [21]

Answer:

2

Step-by-step explanation:

15+12=27

54 / 27 = 2

8 0
3 years ago
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