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3 years ago

Graphing the function please help

Mathematics

1 answer:

Anvisha [2.4K]3 years ago

7 0

Answer:

Step-by-step explanation:

In the function y = -2x -2, the -2x determines where the point will intersect the x-axis and -2 determines what the graph is moving by (going down in this case).

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Adrian spends $238 on a DVD. If he buys 18 DVD,s approximately how much money will he spend in all?

bija089 [108]

Answer:

Adrian would spend exactly $4,284.

Step-by-step explanation:

Multiply the cost by the number of products purchased:

$238 x 18 = $4,284

Have a good day :)

7 0

3 years ago

Read 2 more answers

Help.? geometry basics homework 2: segment addition postulate

Neporo4naja [7]

The value of x is 3 given that ST = 8x + 11 and TU = 12x-1

The point that bisects a line divides the line into two equal parts

If T is the midpoint of SU, the following are true:

ST = TU
ST + TU = SU

Given the following

ST = 8x + 11

TU = 12x-1

Since ST = TU

8x+11 = 12x-1

8x - 12x = -1 - 11

-4x = -12

x = 12/4

x = 3

Hence the value of x is 3 given that ST = 8x + 11 and TU = 12x-1

Learn more here: brainly.com/question/17204733

5 0

2 years ago

Read 2 more answers

Drag and drop the words below to correctly complete the statement of the SAS Similarity Theorem.

Olegator [25]

Congruent
An angle
Sides
Proportional
Similar

5 0

3 years ago

A lottery ticket has a grand prize of $41.7 million. The probability of winning the grand prize is .000000023. Determine the exp

Sonja [21]

Answer:

000023 because if we take three 0's we can get 000023 and 417 million

3 0

2 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