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4 years ago

How do I do this problem

Mathematics

2 answers:

melisa1 [442]4 years ago

6 0

What your going to do is a keep change flip so you keep 1/2, change the division sign to a multiplication sign and flip the 14/5 into 5/14 after that you multiply straight across

sukhopar [10]4 years ago

5 0

1/2÷14/5
1/2x5/14
5/28

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What is the solution to this inequality?

Snezhnost [94]

Answer:

$\boxed{\sf{D. \quad x < 3}}$

Step-by-step explanation:

It is necessary to isolate the term of x from one side of an inequality to find its solution.

First, add by 8 from both sides.

$\sf{-4x-8+8 > -20+8}$

Solve.

-20+8=-12

Rewrite the problem down.

$\sf{-4x > -12}$

Multiply by -1 from both sides.

(-4x)(-1)<(-12)(-1)

Solve.

Multiply the numbers from left to right.

(-12)(-1)=12

4x<12

Divide by 4 from both sides.

4x/4<12/4

Solve.

Divide the numbers from left to right.

12/4=3

x<3

Therefore, the correct answer is D. x<3 (last option).

I hope this helps you! Let me know if my answer is wrong or not.

5 0

3 years ago

Read 2 more answers

Tickets that cost $88 for a professional women’s championship soccer game are offered at a 15% discount. What is the sale price?

RoseWind [281]

Answer:

$74.80

Step-by-step explanation:

10% = 8.8

5% = 4.4

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88 - 13.2 =$74.80

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.