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3 years ago

The greatest common factor 40j-16=

Mathematics

1 answer:

muminat3 years ago

4 0

Answer:

The greatest common factor is 8. So your answer would be: 8(5j - 2) because 8 x 5j= 40j and 8 x 2= 16

Step-by-step explanation:

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RSB [31]

Answer:

hmmmmm

Step-by-step explanation:

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3 years ago

Rectangle A measures 9 in by 3 in in rectangle B is a scaled copy of rectangle A. all measurements

taurus [48]

Answer:
3
Step-by-step explanation:
2 +2 =4 -1 is 3

5 0

3 years ago

Tell whether the given value is a solution of the inequality.<br> x + 1 > 10; x = 9

Oliga [24]

Answer:

given x+1 > 10,

x > 10 - 1

x > 9

which means x MUST BE more than 9 , which also means 9 isnt included in the inequality therefore x = 9 is not a solution.

6 0

3 years ago

1. Write the form of the partial fraction decomposition of the rational expression. Do not solve for the constants.

german

Answer:

Step-by-step explanation:

To write the form of the partial fraction decomposition of the rational expression:

We have:

$\mathbf{\dfrac{8x-4}{x(x^2+1)^2}= \dfrac{A}{x}+\dfrac{Bx+C}{x^2+1}+\dfrac{Dx+E}{(x^2+1)^2}}$

Using partial fraction decomposition to find the definite integral of:

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}dx$

By using the long division method; we have:

$x^2-8x-20 | \dfrac{2x}{2x^3-16x^2-39x+20 }$

$- 2x^3 -16x^2-40x$

$x+ 20$

So;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x+\dfrac{x+20}{x^2-8x-20}$

By using partial fraction decomposition:

$\dfrac{x+20}{(x-10)(x+2)}= \dfrac{A}{x-10}+\dfrac{B}{x+2}$

$= \dfrac{A(x+2)+B(x-10)}{(x-10)(x+2)}$

x + 20 = A(x + 2) + B(x - 10)

x + 20 = (A + B)x + (2A - 10B)

Now; we have to relate like terms on both sides; we have:

A + B = 1 ; 2A - 10 B = 20

By solvong the expressions above; we have:

$A = \dfrac{5}{2}$ $B = \dfrac{3}{2}$

Now;

$\dfrac{x+20}{(x-10)(x+2)} = \dfrac{5}{2(x-10)} + \dfrac{3}{2(x+2)}$

Thus;

$\dfrac{2x^3-16x^2-39x+20}{x^2-8x-20}= 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)}$

Now; the integral is:

$\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = \int \begin {bmatrix} 2x + \dfrac{5}{2(x-10)}+ \dfrac{3}{2(x+2)} \end {bmatrix} \ dx$

$\mathbf{\int \dfrac{2x^3-16x^2-39x+20}{x^2-8x-20} \ dx = x^2 + \dfrac{5}{2}In | x-10|\dfrac{3}{2} In |x+2|+C}$

3. Due to the fact that the maximum words this text box can contain are 5000 words, we decided to write the solution for question 3 and upload it in an image format.

Please check to the attached image below for the solution to question number 3.

4 0

3 years ago

Help me on this please

erica [24]

9514 1404 393

Answer:

sin(θ) = 4/5
cos(θ) = 3/5
tan(θ) = 4/3

Step-by-step explanation:

The mnemonic SOH CAH TOA was created to help you remember the relevant trig relationships. It is intended to remind you ...

Sin = Opposite/Hypotenuse

sin(θ) = 4/5

Cos = Adjacent/Hypotenuse

cos(θ) = 3/5

Tan = Opposite/Adjacent

tan(θ) = 4/3

In a right triangle, the opposite side (PQ) is the one that is not part of the angle (∠R = θ). The adjacent side (QR) is the shorter side that makes up the acute angle. The longest side (PR) is the hypotenuse.

4 0

3 years ago