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3 years ago

An airplane 30,000 feet above the ground

Mathematics

1 answer:

Amiraneli [1.4K]3 years ago

8 0

Answer:

553

Step-by-step explanation:

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4 years ago

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The radius of a right circular cylinder is given by √(t+6) and its height is 1/6√t , where t is time in seconds and the dimensio

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Answer:

The rate change of volume of the cylinder is $\frac{\pi}{4} ( \sqrt t+\frac2{ \sqrt t})$ cubic inch per second.

Step-by-step explanation:

Given that the radius of right circular cylinder is $\sqrt{(t+6)}$ and its height is $\frac16 \sqrt t$ where t is time in second and the dimension are inches.

$\therefore r = \sqrt{(t+6)}$

The base area of the cylinder is A= $\pi r^2$

$=\pi (\sqrt{t+6})^2$

$= \pi (t+6)$

$\therefore A= \pi(t+6)$

Differentiating with respect to t

$\frac{dA}{dt}=\pi$

$\therefore h=\frac16\sqrt t$

Differentiating with respect to t

$\frac{dh}{dt}=\frac16 \times \frac12(t)^{\frac12-1}$

$\Rightarrow \frac{dh}{dt}=\frac1{12} (t)^{-\frac12}$

The volume of cylinder is V= Ah

∴V= Ah

Differentiating with respect to t

$\frac{dV}{dt}=A\frac{dh}{dt}+h\frac{dA}{dt}$

$=\pi (t+6). \frac1{12}t^{-\frac12} +\frac16\sqrt t . \pi$

$=\pi. \frac1{12}.t^{\frac12}+\pi . 6.\frac1{12} t^{-\frac12} +\pi\frac16 \sqrt t$

$=\frac{\pi}{4} ( \sqrt t+\frac2{ \sqrt t})$

The rate change of volume of the cylinder is $\frac{\pi}{4} ( \sqrt t+\frac2{ \sqrt t})$ cubic inch per second.

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3 years ago

Gene has a gasoline budget of $300 per month. He uses an average of $6 of gasoline each day he drives. Which of the following eq

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Answer:

300-6x

Step-by-step explanation:

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2 years ago

I need help with this division problem 1 divide by 3.0

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Answer:

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Step-by-step explanation:

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3 years ago

M∠CBD=4x+52∘

liq [111]

Answer: The value of ∠CBD is 98°.

Step-by-step explanation:

Since we have given that

m∠CBD=4x+52∘

and

m∠ABC=8x−10∘

As we know that they are linear pairs.

As we know that a linear pair is a pair of adjacent angles when two lines get intersected by a line say BC.

So, Sum of linear pair is supplementary.

$\angle CBD+\angle ABC=180^\circ\\\\4x+52+8x-10=180^\circ\\\\12x+42=180^\circ\\\\12x=180^\circ-42^\circ\\\\12x=138^\circ\\\\x=\frac{138}{12}\\\\x=11.5^\circ$

So, m∠CBD = 4x+52=4(11.5)+52=98°

Hence, the value of ∠CBD is 98°.

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4 years ago

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