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2 years ago

8

Find parametric equations for the sphere centered at the origin and with radius 3. Use the parameters s and t in your answer. x(

s, t) = y(s, t) = , and z(s, t) = , where |

1 answer:

Harlamova29_29 [7]2 years ago

4 0

Answer:

Step-by-step explanation:

In general (x, y, z) = (p ·cos s ·sin t, p· sin s· sin t, p · cos t) where p is radius

x(s, t)= 3 ·cos s ·sin t

y(s, t)= 3· sin s· sin t

z(s, t) = 3· cos t

Where, radius is 3, s ∈ [0,2π), and t ∈ [0, π]

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Could someone please help me with this question

defon

Answer:

Proportional: second and third, non-proportional: first and fourth

Step-by-step explanation:

Proportional relationships are linear functions that pass through the origin. Using this definition, we can conclude that the proportional relationships are the second and third options whereas the non-proportional relationships are the first and fourth options.

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2 years ago

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Step-by-step explanation:

5^2 + b^2 = 9^2

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3 years ago

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How to find increase percent from 44 to 48

lesantik [10]

9.09% increase.

First find the difference between the two numbers.

(4)

then, divide the result by the original amount...

4÷ 44

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3 years ago

Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th

trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

t = 20: 999 fishes

t = 30: 1168 fishes

(c)

$P(\infty) = 1200$

Step-by-step explanation:

Given

$P(t) =\frac{d}{1+ke^-{ct}}$

$d = 1200\\k = 11\\c=0.2$

Solving (a): Fishes at t = 0

This gives:

$P(0) =\frac{1200}{1+11*e^-{0.2*0}}$

$P(0) =\frac{1200}{1+11*e^-{0}}$

$P(0) =\frac{1200}{1+11*1}$

$P(0) =\frac{1200}{1+11}$

$P(0) =\frac{1200}{12}$

$P(0) = 100$

Solving (a): Fishes at t = 10, 20, 30

$t = 10$

$P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483$

$t = 20$

$P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999$

$t = 30$

$P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168$

Solving (c): $\lim_{t \to \infty} P(t)$

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of $e^{-ct}$ decreases

This implies that:

${t \to \infty} = {e^{-ct} \to 0}$

So:

The value of P(t) for large values is:

$P(\infty) = \frac{1200}{1 + 11 * 0}$

$P(\infty) = \frac{1200}{1 + 0}$

$P(\infty) = \frac{1200}{1}$

$P(\infty) = 1200$

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2 years ago

Which of the following sets of ordered pairs represents a function?

allsm [11]

Answer:
A

Hope i helped :)

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2 years ago