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3 years ago

Write the expression i35 in the standard form

Mathematics

1 answer:

Sveta_85 [38]3 years ago

6 0

I'm sorry if I'm wrong but I think the answer is -31
No wait nvm maybe not I'm so sorry I can't remember

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3/4x-2=2/5<br> Rewrite so there is no fractions

jek_recluse [69]

Answer:

15x - 40 = 8

Step-by-step explanation:

3/4x-2=2/5

Multiply each side by 4*5 or 20

20 (3/4x-2=2/5)

15x - 40 = 8

7 0

3 years ago

Which of the following sets of numbers could represent the three sides of a triangle

nirvana33 [79]

Answer:

d 5,15,19

Step-by-step explanation:

The sum of the shortest two sides must be longer than the longest side. That is only the case for selection D.

If the sum is equal to the longest side, the segment ends will meet, but the "triangle" will look like a line segment. Many authors do not allow such a thing to be called a triangle. Hence, 'b' is not an answer to this question.

(Some authors <em>do</em> allow such a triangle, in which case, there would be two answers here: B and D.)

3 0

3 years ago

a can of pinapple juice is a cylinder with a radius of 2.4 in. and a height of 11. what is the are odf the label around the can

pshichka [43]

A=2πrh+2πr²
A≈202.07
hopes that helps

8 0

2 years ago

PLEASE HELP 9 AND 10 <br> IM LITERALLY CRYING

tekilochka [14]

Answer:

#10:

burger: $7.50

fries and drink: 2.50

idk about #9 tho sry

Step-by-step explanation:

6 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