Please i need help ASAP!!!

Cancel the common factor of the numerator and the denominator and write specified expression

lisabon 2012 [21]

Step-by-step explanation:

Hello,

I hope you mean to cancel the common factor that exists in numerator and denominator,right.

so, Let's look for the common factor, 

here, the expression is, 

=4(x-2)/ (x+5)(x-2)

so, here we find the common factor is (x-2)

now, we have to cancel it. And after cancelling we get,

=4/(x+5)

Note:{ we cancel the common factor if the common factors are in multiply form.}

Hope it helps 

3 0

3 years ago

What is the greatest common factor (GCF) of 88 and 132?

maw [93]

I’m pretty sure the gcf is 44

8 0

3 years ago

Read 2 more answers

(Geometry) PART A Find the value of X.

guapka [62]

Answer:

Part A: The answer is B.

As vertically opposite angles are equal

x+10=4x-35

10=3x-35

45=3x

x=15

Part B: The answer is D.

We'll substitute in the value of X in one equation.

x+10

=15+10

=25

As the two angles are equal bot must be 25 degrees.

5 0

2 years ago

What is the length of the curve with parametric equations x = t - cos(t), y = 1 - sin(t) from t = 0 to t = π? (5 points)

zzz [600]

Answer:

B) 4√2

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Parametric Differentiation

Integration

Integrals
Definite Integrals
Integration Constant C

Arc Length Formula [Parametric]: $\displaystyle AL = \int\limits^b_a {\sqrt{[x'(t)]^2 + [y(t)]^2}} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \left \{ {{x = t - cos(t)} \atop {y = 1 - sin(t)}} \right.$

Interval [0, π]

Step 2: Find Arc Length

[Parametrics] Differentiate [Basic Power Rule, Trig Differentiation]: $\displaystyle \left \{ {{x' = 1 + sin(t)} \atop {y' = -cos(t)}} \right.$
Substitute in variables [Arc Length Formula - Parametric]: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{[1 + sin(t)]^2 + [-cos(t)]^2}} \, dx$
[Integrand] Simplify: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx$
[Integral] Evaluate: $\displaystyle AL = \int\limits^{\pi}_0 {\sqrt{2[sin(x) + 1]} \, dx = 4\sqrt{2}$

Topic: AP Calculus BC (Calculus I + II)

Unit: Parametric Integration

Book: College Calculus 10e

4 0

2 years ago

What is the solution of the following system?

Yakvenalex [24]

<h3>Answer: B) no solutions</h3>

===========================================================

Explanation:

Let's isolate the variable y in the first equation

-2x - y = 1

-2x = 1 + y

1+y = -2x

y = -2x-1

Then we'll plug this into the second equation

-4x - 2y = -1

-4x - 2(-2x-1) = -1

-4x + 4x + 2 = -1

0x + 2 = -1

0 + 2 = -1

2 = -1

We run into a contradiction. No matter what x value we pick, the last equation will never be true. This leads to a domino effect to cause the first set of original equations to never be true when considered as a system.

Therefore, this system is inconsistent. There are no solutions. These two lines graph out parallel lines that never intersect.

3 0

2 years ago

Please i need help ASAP!!!​

Please i need help ASAP!!!