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valentinak56 [21]
3 years ago
7

Solve a distributive property equation algebraically. Show all work and all steps.

Mathematics
1 answer:
boyakko [2]3 years ago
5 0
6(3a-2)+5a=57

18a-12+5a=57

23a-12=57

23a=69

a=3


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Evaluate the integral:<br> dx / e^(pie*x) from 2 (upper bound) to 0.
kenny6666 [7]
Substitution:
u = - π x
d u = - π d x
d x = - d u / π 
----------------------
-1/ \pi  \int\limits^2_0 {e ^{u} } \, du =- \frac{e ^{- \pi x} }{ \pi } = \\ -1/ \pi (e ^{-2 \pi } -1) = \\ = \frac{1}{ \pi } - \frac{1}{ \pi e ^{2 \pi } }
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3 years ago
Gabriella swims 2 laps per minute. What is the associated rate? ___ minute per lap
Leokris [45]

Answer:

The answer is D.

Step-by-step explanation:

It is given that Gabriella swims 2 laps for 1 minute. So in order to find how many minutes did he swim for each lap, you have to divide it by 2 :

2 laps = 1 minute

2 laps ÷ 2 = 1 minute ÷ 2

1 lap = 1/2 minute

= 30 seconds

Test :

Gabriella swim 1/2 min per lap. So if he swim 2 laps, you have to multiply it by 2 :

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7 0
3 years ago
Read 2 more answers
ben has 3 cookies for each guest plus a 5 extra cookies for each guest write a word problem to solve this
quester [9]

Answer: Three plus Five equals Eight or 3= a · 5= 5a

Step-by-step explanation:

7 0
3 years ago
Dang it these are tricky for me can somebody help?
erica [24]

Answer:

It has no perfect square factors, unless you count 1 = 1^2. That's sort of a degenerate case we don't usually count, since every integer has that factor. We would usually say that 1290 is a square-free integer. Hope this helps!!

5 0
4 years ago
Solving compound inequalities <br><br> 14r+20 &lt;14r+16 or 8-10r&gt;15-9r
vova2212 [387]
14r+20<14r+16 or 8−10r>15−9r

14r+20<14r+16

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20<16

20−20<16−20(Subtract 20 from both sides)

0<−4



8−10r>15−9r

−10r+8>−9r+15(Simplify both sides of the inequality)

−10r+8+9r>−9r+15+9r(Add 9r to both sides)

−r+8>15

−r+8−8>15−8(Subtract 8 from both sides)

−r>7

−r/−1 > 7/−1(Divide both sides by -1)

r<−7




Ur answer is:

r < - 7  OR   0 < - 4

7 0
3 years ago
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