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3 years ago

PLS HELP ASAP Function g is shown on the graph Which function represents g

Mathematics

1 answer:

Marat540 [252]3 years ago

3 0

Answer:

Step-by-step explanation:

A piecewise function g(x) is represented by the graph. Which functions represent a piece of the function? Check all that apply.

g(x) = −2x, −2 < x < 0

g(x) = −2, x < −2

g(x) = x − 2, −2 < x < 1

g(x) = −2x + 6, x ≥ 1

g(x) = x/2+ 1, –2 ≤ x < 1

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Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

A sphere is inscribed in a cube with a volume of 64 cubic inches. What is the volume of the sphere? Round your answer to the nea

erma4kov [3.2K]

Answer:

the volume of the sphere is

$33.51 in^{3}$

Step-by-step explanation:

This problem bothers on the mensuration of solid shapes, sphere and cube.

Given data

Volume of cube v = 64 cubic inches

since we are dealing with a cube the height and the radius of the sphere is same as the sides of the cube,

we know that volume of cube is expressed as

$v= l*b*h$

$v=l^{3}$

$64= l^{3}$

$l= \sqrt[3]{64}$

$l= 4 in$

also diameter d=length l

Diameter d= $4in$

Radius r = $\frac{d}{2}$ = $\frac{4}{2}$ = ${2 in}$

Height h= $4in$

we know that the volume of a sphere is given by

$v= \frac{4}{3} \pi r^{3}$

substituting into the formula we have

$v= \frac{4}{3} \ *3.142*2^{3} \\v=\frac{4*3.142*8}3} \\v= \frac{100.54}{3} \\v= 33.52in^{3}$

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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,

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

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

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3 years ago