Ask question

Ask question

All categories

3 years ago

12

Simplify the rational Expression

1 answer:

qaws [65]3 years ago

4 0

Answer:

$\frac{x^2+2x}{x-3}$

Step-by-step explanation:

We need to factor out numerator and denominator in order to simplify the rational expression by cancelling common factors.

Numerator : x^3 - 4 x = x (x^2 - 4) = x (x - 2) (x + 2)

Denominator (factoring by grouping):

x^2 - 5 x + 6 = x^2 - 3 x - 2 x + 6 = x (x - 3) - 2 (x - 3) = (x - 3) (x - 2)

Then we can cancel out the common factor (x - 2) in both numerator and denominator, leading to:

x (x + 2) / (x - 3) = (x^2 + 2)/ (x-3)

$\frac{x^2+2x}{x-3}$

You might be interested in

How to get roots from quadratic formula

jeyben [28]

Answer:

factoring / standard form

6 0

4 years ago

Assume that a procedure yields a binomial distribution with n=3 trials and a probability of success of p=0.300. use a binomial p

NNADVOKAT [17]

$\mathbb P(X=x)=\begin{cases}\dbinom3xp^x(1-p)^{3-x}&\text{for }0\le x\le3\\\\0&\text{otherwise}\end{cases}$

$\implies\mathbb P(X=2)=\dbinom32(0.3)^2(0.7)^1=0.189=18.9\%$

7 0

4 years ago

In ΔQRS, q = 4.9 cm, r = 7.1 cm and ∠S=70°. Find the area of ΔQRS, to the nearest 10th of a square centimeter.

IgorC [24]

Answer:16.3

Step-by-step explanation:

Delta math

4 0

3 years ago

Please help me answer this question

stealth61 [152]

The value of dy/dx for the functions are

(i) $\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

<h3>Differentiation</h3>

From the question, we are to determine dy/dx for the given functions

(i) $x^{2} sin^{2}2x$

Let $u = x^{2}$

and

$v = sin^{2} 2x$

Also,

Let $w= sin2x$

∴ $v = w^{2}$

First, we will determine $\frac{dv}{dx}$

Using the Chain rule
$\frac{dv}{dx} = \frac{dv}{dw}.\frac{dw}{dx}$

$v = w^{2}$

∴ $\frac{dv}{dw} =2w$

Also,

$w= sin2x$

∴ $\frac{dw}{dx} =2cos2x$

Thus,

$\frac{dv}{dx} = 2w \times 2cos2x$

$\frac{dv}{dx} = 2sin2x \times 2cos2x$

$\frac{dv}{dx} = 4sin2x . cos2x$

Now, using the product rule

$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$

From above

$u = x^{2}$

∴ $\frac{du}{dx}=2x$

Now,

$\frac{dy}{dx} = x^{2} (4sin2x.cos2x)+ sin^{2}2x (2x)$

$\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $xy^{2}+y^{2}x^{3} +2=0$

Then,

$x.2y\frac{dy}{dx}+ y^{2}(1)+y^{2}.3x^{2} + x^{3}.2y\frac{dy}{dx} +0=0$

$2xy\frac{dy}{dx}+ y^{2}+3x^{2}y^{2} + 2x^{3}y\frac{dy}{dx} =0$

$2xy\frac{dy}{dx}+2x^{3}y\frac{dy}{dx} =- y^{2}-3x^{2}y^{2}$

$\frac{dy}{dx} (2xy+2x^{3}y)=- y^{2}(1+3x^{2})$

$\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy+2x^{3}y}$

$\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy(1+x^{2}) }$

$\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

Hence, the value of dy/dx for the functions are

(i) $\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x$

(ii) $\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }$

Learn more on Differentiation here: brainly.com/question/24024883

#SPJ1

8 0

2 years ago

Brittany will be working full time this summer to save for her goal of having $10,000 by the time shes 21. Brittany has an accou

Scrat [10]

Answer:

Guys i cant even readhhdhshhsjsvfjdjznbx

7 0

3 years ago