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3 years ago

Three quarters are what fraction of a dollar

Mathematics

2 answers:

barxatty [35]3 years ago

8 0

Three quarters are 3/4 of a dollar

Nonamiya [84]3 years ago

5 0

Three quarters are 3/4 of a dollar.

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Define a variable and write an equation for each real-world problem. 1] Matt spent $ 6.50 on his lunch. This is $2.25 more than

sleet_krkn [62]

M will represent Matt and J is Jackson. M-2.25=J. You minus the difference between them and show that it equals to Jacksons. A variable is a number in math that can vary.

8 0

3 years ago

3. The volume of a book is 60 cubic inches. If a large

saveliy_v [14]

Answer:

720 in^3.

Step-by-step explanation:

12 * 60 = 720.

6 0

2 years ago

Read 2 more answers

What is the value of f(g(1))

jeka57 [31]

Answer: 1

Step-by-step explanation:

All you had to do was multiply g by 1 so f(g(1)) as a value is 1

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

Which one is the correct answer?

Gala2k [10]

The second one is the right answer

7 0

3 years ago

Read 2 more answers