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Varvara68 [4.7K]

3 years ago

8

Factor completely 6w^4 – 54w^2

2 answers:

kondaur [170]3 years ago

5 0

$6w^4-54w^2=6w^2\cdot w^2-6w^2\cdot9=6w^2\cdot(w^2-9)=6w^2(w-3)(w+3)$

MaRussiya [10]3 years ago

4 0

$6w^4-54w^2=6w^2(w^2-9)=6w^2(w^2-3^2)=6w^2(w-3)(w+3)$

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10 Points! Help is aprreciated!! :)

Ugo [173]

Answer:

A

Step-by-step explanation:

Since the the value is greater than 1, f(x),g(x), and h(x) are exponential functions.

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4 years ago

Consider the function f(x)=x3+15x2+74x+120. If f(x)=0 for x=−6, for what other values of x is the function equal to 0? List the

jasenka [17]

Answer:

other x values are -5,-4

Step-by-step explanation:

f(x)=0 for x=−6

Apply synthetic division to get the quotient . using that we find other two values of x

-6 1 15 74 120

0 -6 - 54 -120

-------------------------------------------------

1 9 20 0

the quotient is $x^2+9x+20=0$

now factor the left hand side, product is 20 and sum is 9

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So other x values are -5,-4

8 0

3 years ago

When simplifying rational numbers, what would the answer be to 57/69

Irina-Kira [14]

You don't simplify rational numbers, but sometimes you can simplify a fraction.
The way to do that is to divide the top and bottom by their greatest common factor.

The greatest common factor of 57 and 69 is 3 . When you divide the top and
bottom by 3 , you have 19/23 . That's the simplest form of that particular fraction.

5 0

3 years ago

Read 2 more answers

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

An equation of the line that passes through the given point and is parallel to the graph of the given equation

notsponge [240]

To construct the equations for these lines, we are going to use the point-slope formula, which is:

$(y - y_1) = m(x - x_1)$

$m$ is the slope of the line
$(x_1, y_1)$ is a point on the line

1)

In this case, our point is (2, -1) and our line has a slope of 5. (Remember that the line is parallel, meaning that it has the same slope as the given line. Since the line is in the y = mx + b format, we could easily pick out the slope m as being 5.)

Thus, we can "plug in" what we know into our formula to find the equation of our line:

$(y + 1) = 5(x - 2)$

$(y + 1) = 5x - 10$

$y = 5x - 11$

2)

We are going to do the same thing, except our point is now (0, -5) and our slope is now 9.

$(y + 5) = 9(x - 0)$

$y = 9x - 5$

Our equations are:

1) y = 5x - 11

2) y = 9x - 5

7 0

3 years ago