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3 years ago

10

I need help with another set of math questions.

1 answer:

Anit [1.1K]3 years ago

7 0

For question 11, you essentially need to find when h(t) = 0, since that is when the height of the ball reaches 0 (ie touches the ground).

For question 12, it is asking for a maximum height, so you need to find when dh/dt = 0 and taking the second derivative to prove that there is maximum at t. That will find you the time at which the ball will hit a maximum height.

Rinse and repeat question 12 for question 13

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allochka39001 [22]

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Your answer is D^

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romanna [79]

Answer:

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Step-by-step explanation:

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Sveta_85 [38]

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Step-by-step explanation:

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3 years ago

Among all right circular cones with a slant height of 24, what are the dimensions (radius and height) that maximize the volum

aleksklad [387]

Answer:

5571.99

Step-by-step explanation:

We need to use the Pythagorean theorem to solve the problem.

The theorem indicates that,

$r^2+h^2=24^2 \\r^2+h^2=576\\r^2=576-h^2$

Once this is defined, we proceed to define the volume of a cone,

$v=\frac{1}{3}\pi r^2 h$

Substituting,

$v=\frac{1}{3} \pi (576-h^2)h\\v=\frac{1}{3} \pi (576h-h^3)$

We need to find the maximum height, so we proceed to calculate h, by means of its derivative and equalizing 0,

$\frac{dv}{dh} = \frac{1}{3} \pi (576-3h^2)$

$\frac{dv}{dh} = 0$ then $\rightarrow \frac{1}{3}\pi(576-3h^2)=0$

$h_1=-8\sqrt{3}\\h_2=8\sqrt{3}$

<em>We select the positiv value.</em>

We have then,

$r^2 = 576-(8\sqrt3)^2 = 384\\r=\sqrt{384}$

We can now calculate the maximum volume,

$V_{max}= \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (\sqrt{384})^2 (8\sqrt{3}) = 5571.99$

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3 years ago

Find the surface area of cube whose volume is 64 cm cube

Oksana_A [137]

V = L³ = 64 => L = ∛64 = 4 cm

A = L² = 4² = 16 cm²

At = 16 x 6 = 96 cm²

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3 years ago