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2 years ago

Math #lollololollollol

Mathematics

2 answers:

Gekata [30.6K]2 years ago

6 0

Answer: its the second one (2)

Ivahew [28]2 years ago

4 0

Answer:

the answer is 2

Step-by-step explanation:

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The graph of y=f(x) is above. If f(−4)=k, what is the value of f(k)?

cupoosta [38]

Answer:

The answer is B, which is -1.5.

3 0

3 years ago

FAST PLSSSSS

hodyreva [135]

Answer:

factor:

x^2+2x+2x+4

x(x+2)+2(x+2)

(x+2)(x+2)

100x^2+20x+1

100x^2+10x+(10x+1)

10(10x+1)+(10x+1)

(10x+1)(10x+1)=(x+2)(x+2)

Hope this helps you

6 0

2 years ago

Read 2 more answers

I will give brainliest if correct

dalvyx [7]

Answer:supplementary

Step-by-step explanation:

Opposite angles in any quadrilateral inscribed in a circle are supplements of each other.

5 0

3 years ago

Make 'y' the subject of the equation. x = 2y -7

geniusboy [140]

Answer:

y = (x + 7)/2

here's your solution

=>. it is given. x = 2y - 7

=> we have to make y as subject

=> now, we can writ this equation in this way

=> 2y - 7 = x

=> now put all constant to right side

=> 2y = x +7

=>. y = x+ 7 /2

hope it helps

5 0

2 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

2 years ago