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3 years ago

10

What is the value of 5.7 x 10^8?

2 answers:

aniked [119]3 years ago

6 0

Its d 570,000,000 hope it helps

stealth61 [152]3 years ago

4 0

Answer:

D

Step-by-step explanation:

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Toss coin 3 times. What is probability of getting 1 head and 2 tails in any order?

Ugo [173]

1/8 because

1/2 x 1/2 x 1/2

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3 years ago

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The point A(-7, 3) is reflected over the point (-4, 5) and its image is point B.

Rudiy27

(7, 3) is the answer I believe

3 0

2 years ago

How many different twotwo-letter passwords can be formed from the letters

Anna [14]

Answer:

$70$

Step-by-step explanation:

GIVEN: two two-letter passwords can be formed from the letters A, B, C, D, E, F, G and H.

TO FIND: How many different two two-letter passwords can be formed if no repetition of letters is allowed.

SOLUTION:

Total number of different letters $=8$

for two two-letter passwords $4$ different are required.

Number of ways of selecting $4$ different letters from $8$ letters $=^8C_4$

$=\frac{8!}{4!4!}$

$=70$

Hence there are $70$ different two-letter passwords can be formed using $8$ letters.

5 0

3 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

Struggling!!! Math help plss!!!

Lina20 [59]

Answer:

Step-by-step explanation:

3 0

2 years ago