Math Question Please Help

Determine whether the given value is a solution of the equation x over nine=4; x=36

Citrus2011 [14]

Yes ; "x = 36" , is, in fact, a solution.
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Given the equation: "x / 9 = 4" ;
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"x = 36" is a solution ; since "36 ÷ 9 = 4" ; and since x = (4)*(9) = 36.
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8 0

4 years ago

Read 2 more answers

Joseph's lunch at a restaurant costs 13.00, without tax. He leaves the waiter a tip of 17% of the cost of the lunch, without tax

Musya8 [376]

Answer:

The cost before tax with tip is $15.21.

Step-by-step explanation:

To calculate percentage, take the percent (17%) and take its decimal form. This means 17% is 0.17 (like 50% would be 0.5). multiply the number you want the percentage of by the decimal form. 13*0.17=2.21. Add this to the cost before tax and before tip. 13.00+2.21=15.21. The cost before tax with tip is $15.21.

3 0

3 years ago

If r/s = 22 and the value of r is 242, what is the value of s

jekas [21]

Answer:

s = 11

Step-by-step explanation:

Given

$\frac{r}{s}$ = 22 ← substitute r = 242

$\frac{242}{s}$ = 22 ( multiply both sides by s )

242 = 22s ( divide both sides by 22 )

11 = s

7 0

3 years ago

Read 2 more answers

PLEASE HELP I’LL MAKE YOU THE BRAINIEST <br> find the area of the trapezoid to the nearest tenth

murzikaleks [220]

Answer:

2.7m

Step-by-step explanation:

1.0m

0.5m

+1.2m

2.7m - answer

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:

1.

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

We have:

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

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