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3 years ago

5

Marcus said the product of 7/8 × 1 3/4 is 9/4. How can you tell that his answer is wrong?

1 answer:

Paraphin [41]3 years ago

8 0

Gravity simply does not allow that

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6. Which of the following is true about the

Gnom [1K]

C.
Y=mx + b is the formula (M being slope, and B being the y intercept) always start by going up or down by what the y-intercept number is, if it’s -9 you go down the y axis 9 numbers from the origin (0,0) and slope (m) is always better put in a fraction-5x is better put in -5/1x
Think of the top number of the fraction as ‘up/down’ and the bottom number as ‘over’ if the top number is negative you will go down from the y-intercept number that many times, the bottom number is how many spots you go over from that number , it is always to the right.

5 0

3 years ago

1-3x=8x+3 solve the equation

LenaWriter [7]

First, subtract both sides by 8x and you would get:

1 - 11x = 3

Then, subtract both sides by 1 and you will get:

-11x = 2

Divide both sides by -11:

x = -2/11

8 0

3 years ago

If the area of a square is 32ft^2 how long is its diagonal

denpristay [2]

Answer:

45.25

Step-by-step explanation:

32^2+32^2=2048 sqrt2048 is 45.25

7 0

3 years ago

Read 2 more answers

HELP PLS ASAP WILL GIVE 5 STARS

Andreyy89

A

Step-by-step explanation:

I do now know what this is about

7 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

3 years ago