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2 years ago

7

Divide this polynomial please thank you

1 answer:

Novosadov [1.4K]2 years ago

7 0

Answer:

Using polynomial division, the answer would come out to be:

3x^2-4x-4+(1/2x-1)

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If you had to simplify 15/24 what would it be

Effectus [21]

If I had to simplify 15/24 it would be 5/8

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3 years ago

Read 2 more answers

Rewrite the expression as an algebraic expression in x. <br> cos(tan−1 x)

zhenek [66]

It's hard to type and hard to read the "inverse tangent" function, as you've seen (above).

So, use "arctan x" instead.

Then the problem becomes: "differentiate cos (arctan x)."

You must apply first the rule for differentiating the cosine function, and next apply the rule for differentiating the arctan function:

(d/dx) cos (arctan x) = - sin (arctan x) * [1/(1+x^2)]

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3 years ago

HELP ME PLAESE ASAP PLAESE TELL ME HOW YOU SOLVED ITT

dolphi86 [110]

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3 years ago

Find gradient <br><br>xe^y + 4 ln y = x² at (1, 1)

cricket20 [7]

$xe^y+4\ln y=x^2$

Differentiate both sides with respect to <em>x</em>, assuming <em>y</em> = <em>y</em>(<em>x</em>).

$\dfrac{\mathrm d(xe^y+4\ln y)}{\mathrm dx}=\dfrac{\mathrm d(x^2)}{\mathrm dx}$

$\dfrac{\mathrm d(xe^y)}{\mathrm dx}+\dfrac{\mathrm d(4\ln y)}{\mathrm dx}=2x$

$\dfrac{\mathrm d(x)}{\mathrm dx}e^y+x\dfrac{\mathrm d(e^y)}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x$

$e^y+xe^y\dfrac{\mathrm dy}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x$

Solve for d<em>y</em>/d<em>x</em> :

$e^y+\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x$

$\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x-e^y$

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x-e^y}{xe^y+\frac4y}$

If <em>y</em> ≠ 0, we can write

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2xy-ye^y}{xye^y+4}$

At the point (1, 1), the derivative is

$\dfrac{\mathrm dy}{\mathrm dx}\bigg|_{x=1,y=1}=\boxed{\dfrac{2-e}{e+4}}$

4 0

3 years ago

assume x and y are both differentiable functions of t. find dx/dt given x=-1 and dy/dt=8 for the relation: 4x^2+3y^3=28

melomori [17]

Given:

x and y are both differentiable functions of t.

$4x^2+3y^3=28$

$x=-1\text{ and }\dfrac{dy}{dt}=8$

To find:

The value of $\dfrac{dx}{dt}$ .

Solution:

We have,

$4x^2+3y^3=28$ ...(i)

At x=-1,

$4(-1)^2+3y^3=28$

$4+3y^3=28$

$3y^3=28-4$

$3y^3=24$

Divide both sides by 3.

$y^3=8$

Taking cube root on both sides.

$y=2$

So, y=2 at x=-1.

Differentiate (i) with respect to t.

$4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0$

Putting x=-1, y=2 and $\dfrac{dy}{dt}=8$ , we get

$4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0$

$-8\dfrac{dx}{dt}+9(4)(8)=0$

$-8(\dfrac{dx}{dt}-9(4))=0$

Divide both sides by -8.

$\dfrac{dx}{dt}-36=0$

$\dfrac{dx}{dt}=36$

Therefore, the value of $4x^2+3y^3=28$ is 36.

6 0

2 years ago