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2 years ago

The surface area of 2 spheres are in ration of 1:16 what is the ratio of their volume

Mathematics

1 answer:

crimeas [40]2 years ago

4 0

S1/S2 = 1/16
Thus R1/R2 = 1/4
So V1/V2 = 1/64
Answer: 1/64

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When does it reach maximum height?<br><br> H(t)=-16t^2+112t+288

ivolga24 [154]

Answer:

484 feet maximum

Step-by-step explanation:

H(t)=-16t^2+112t+288

The maximum height occurs at the vertex of this graph.

Let us find the vertex.

vertex x-coordinate = -b/2a

x = -b/2a = -(112)/ (2*-16) = 7/2 sec.

H(7/2) = - 16 *(7/2)^2 + 112*(7/2) + 288

H(7/2) = -196 + 392 + 288 = 484 feet ..

5 0

2 years ago

SUPER EASY QUESTION!!! What is 6.3 rounded to the nearest tenth?

AnnZ [28]

The answer is 6-------------------------

5 0

3 years ago

PLEASE DONT SEND LINKS OR I WILL REPORT! Emily buys meat for a charity event in her community. Chicken costs 4 AED per kg and be

lyudmila [28]

Answer:

21 AED

Step-by-step explanation:

W know the cost of each kind of meat

Chicken = 4 AED

Beef = 3 AED

We also know how much of each we will be buying

Chicken = 3.6 kg

Beef = 2.2 kg

Multiply each price of meat by its respective weight, then add the products together for you final answer.

4(3.6) + 3(2.2) = 21

5 0

2 years ago

Help me with differentation and integration please!!

Marina86 [1]

Answer:

See below

Step-by-step explanation:

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

Recall

$\dfrac{d}{dx}\tan x=\sec^2$

Using the chain rule

$\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}$

such that $u = \tan x$

we can get a general formulation for

$y = \tan^n x$

Considering the power rule

$\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}$

we have

$\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x$

therefore,

$\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x$

Now, once

$\sec^2 x - 1= \tan^2x$

we have

$3\tan^2x \sec^2x = 3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x$

Hence, we showed

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

================================================

For the integration,

$\int \sec^4 x\, dx$

considering the previous part, we will use the identity

$\boxed{\sec^2 x - 1= \tan^2x}$

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx$

Considering $u = \tan x$

and then $du=\sec^2x\ dx$

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0

2 years ago

Is -4 1/5 less than. greater than or equal to -4 9/10

Natalija [7]

-4 1/5 is greater than -4 9/10

8 0

3 years ago