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3 years ago

Simplify. 6(2y +8) -2(3y -2)

Mathematics

2 answers:

Advocard [28]3 years ago

7 0

Answer:

6y+52.

Step-by-step explanation:

What you have to do is the Distributive Property and distribute the 6 and -2 into the parenthesis.

6(2y+8)-2(3y-2).

12y+48-6y+4. You will then add or subtract the same together to get.

6y+52 as your answer.

azamat3 years ago

4 0

Answer:

6y + 52

Step-by-step explanation:

(12y + 48 ) - (6y - 4)

6y + 52

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If the mean of 25, 28, a, 30 and 32 is 27, then find the value of a.

Anna11 [10]

Answer:

a = 20

Step-by-step explanation:

The mean is calculated as

mean = $\frac{sum}{count}$

= $\frac{25+28+a+30+32}{5}$ = $\frac{115+a}{5}$

Then

$\frac{115+a}{5}$ = 27 ( multiply both sides by 5 to clear the fraction )

115 + a = 135 ( subtract 115 from both sides )

a = 20

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3 years ago

Use the distance formula to find the length of AB. Point A (0 , -8) and Point B (3 , -2). Round your answer to the tenths place.

Lana71 [14]

Answer:

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Step-by-step explanation:

7 0

3 years ago

If 17,000 is 47% what is 100%<br> Please explain!

Alja [10]

Answer:

36,170.21276595745 (I did this on a calculator.)

Step-by-step explanation:

17000/47 = 361.7021276595745

361.7021276595745 x 100 = 36,170.21276595745

Your total answer is 36,170.21276595745.

8 0

3 years ago

If ∠A and ∠B are supplementary and m∠A = 37°45', then m∠B = ______.

xenn [34]

Answer: 142°15'

Explanation:

Supplementary angles are angles that add up to make a straight angle ( $180^{\circ}$ ).

Therefore, we must find the angle ∠B such that

∠A + ∠B = 180° (1)

We know that ∠A = 37°45', so we can re-arrange equation (1) to find the magnitude of ∠B:

∠B = 180° - ∠A = 180° - 37°45' = 142°15'

So, the correct answer is

142°15'

6 0

3 years ago

Read 2 more answers

Find a compact form for generating functions of the sequence 1, 8,27,... , k^3

pantera1 [17]

This sequence has generating function

$F(x)=\displaystyle\sum_{k\ge0}k^3x^k$

(if we include $k=0$ for a moment)

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k$

$\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k$

Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

$\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k$

Take the derivative one more time:

$\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k$

$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

5 0

3 years ago

Simplify. 6(2y +8) -2(3y -2)​

Simplify. 6(2y +8) -2(3y -2)