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3 years ago

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The volume of a sphere is 500 3 πcm3. What is the radius? Sphere V = 4 3 πr3 1. Substitute value into formula: 500 3 π = 4 3 πr

3 2. Undo multiplication of 4 3 : 125π = πr3 3. Undo multiplication of π: 125 = r3 4. Undo cube: 3√125 = r The radius of the sphere is 6 ⇒ 5 cm

2 answers:

Soloha48 [4]3 years ago

8 0

Answer:

5m

Step-by-step explanation:

Kisachek [45]3 years ago

7 0

Answer:

Radius of the sphere is 5m.

Step-by-step explanation:

We have given the volume of sphere $\frac{500}{3}\cdot {\pi}m^3$

we have to find the radius of the sphere

$\text{volume of sphere is }=\frac{4}{3}\cdot {\pi}\cdot r^3$

On substituting the given value of volume of sphere we get

$\frac{500}{3}\cdot {\pi}=\frac{4}{3}\cdot {\pi}\cdot r^3$

After simplification we get:

$\Rightarrow 4\cdot r^3=500$

$\Rightarrow r^3=125$

$\Rightarrow r=5$

Hence, the radius of sphere is 5m

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Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

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Answer:

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b) $0$

c) $a(t) = x''(t) = v'(t) =6t-12$

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And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

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Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

$v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

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Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

$v(t) = 3t^2 -12t +9$

We can factorize this function as $v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)$

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is $0\leq t\leq 10$ we need to analyze the following intervals:

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$v(0.25) =6.1875, v(0.5)=3.75$

And we see that for $a=0.5 >0.25=b$ we have that f(b)>f(a) so then the function is decreasing on this case.

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And at t=-2 v(2) = -3 that represent the minimum for this function, we see that if we select two values let's say 1.5 and 1.75

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$2$

We see that the function would be increasing.

$3$

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Part c

$a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

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And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

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The first three work correctly, each one with a 92% probability.

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