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satela [25.4K]
2 years ago
13

Write an equivalent expression in word form. (1- 1/3) / 2

Mathematics
2 answers:
Oksi-84 [34.3K]2 years ago
8 0
One third taken away from one, halved
or
half of one minus one third
KonstantinChe [14]2 years ago
3 0

Answer:

21

Step-by-step explanation:

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D=dollars earned
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(1,5) (2,10) (3,15)

D=5T
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Phil is making a boat. His small model is 24 inches long and 4.75 inches wide. He is finalising it, and it is 8 feet long. How w
Maslowich

your answer would be 19 inches because 24 divided by 12(length of a foot in inches) is 2 and therefore you also divide 4.75 by 2 to get 2.375 and multiply that by 8 the amount of feet and your done

6 0
3 years ago
Solve for C.<br> -5/8c=20
iris [78.8K]
-5/8c = 20....multiply both sides by -8/5 (this cancels out the -5/8 on the left
c = 20(-8/5)
c = -160/5
c = - 32

check..
-5/8(-32) = 20
160/8 = 20
20 = 20 (correct)

so c = -32
8 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
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