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Mademuasel [1]
3 years ago
6

Simplify the following expression: 6(3x) − 2y + 3z + 12(4x) − 9y

Mathematics
1 answer:
Slav-nsk [51]3 years ago
4 0

Answer:

9x+48x-2y-9y +3z

57x -11y +3z

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There are 32 students in the 6th grade at Garden Valley Middle School. Eight of them have freckles. What is the ratio of 6th gra
aleksandr82 [10.1K]

Answer:

A. 1:4 is the answer of this question

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3 years ago
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The length of a rectangle is 5 inches less than twice its width. The perimeter of the rectangle is 26 inches. What is the width
Yakvenalex [24]

Answer:

w= 6

Step-by-step explanation:

I just started by making an educated guess using the values already given. Then I inserted that into the problem to see if it worked.

L= 2w - 5

I used 6 as a random, educated guess for the value of w.

L = 2(6) - 5

L = 12-5

L = 7

Then, multiply L by 2 to account for both side lengths of the rectangle.

7(2)= 14

Subtract that value from the total perimeter to find what the width must equal.

26 - 14 = 12

Divide that answer by 2 since there are two sides for width.

12/2 = 6

I know this was kind of long, but I hope it helps! :)

3 0
3 years ago
Factor the Difference of Square.<br><br> 25x2 _ 16
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3 years ago
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Suppose Eric runs for 4 minutes and walks for 8 minutes how close is he to 1,600 meter goal?
jarptica [38.1K]
Hi there,

Since Eric ran 200 m/min for 4 minutes, then 4×200 = 800m
and he walked 80 m/min for 8 minutes, then 8×80 = 640m
and in total he ran 800 + 640 = 1440m 
so he still have 1600 - 1440 = 160m left to reach 1600m

Hope this helped :P



3 0
4 years ago
Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges.
Arisa [49]

Answer: S_n=5(1-\dfrac{1}{n+1}) ; 5

Step-by-step explanation:

Given series : [\dfrac{5}{1\cdot2}]+[\dfrac{5}{2\cdot3}]+[\dfrac{5}{3\cdot4}]+....+[\dfrac{5}{n\cdot(n+1)}]

Sum of series = S_n=\sum^{\infty}_{1}\ [\dfrac{5}{n\cdot(n+1)}]=5[\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}]

Consider \dfrac{1}{n\cdot(n+1)}=\dfrac{n+1-n}{n(n+1)}

=\dfrac{1}{n}-\dfrac{1}{n+1}

⇒ S_n=5\sum^{\infty}_{1}\dfrac{1}{n\cdot(n+1)}=5\sum^{\infty}_{1}[\dfrac{1}{n}-\dfrac{1}{n+1}]

Put values of n= 1,2,3,4,5,.....n

⇒ S_n=5(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......-\dfrac{1}{n}+\dfrac{1}{n}-\dfrac{1}{n+1})

All terms get cancel but First and last terms left behind.

⇒ S_n=5(1-\dfrac{1}{n+1})

Formula for the nth partial sum of the series :

S_n=5(1-\dfrac{1}{n+1})

Also, \lim_{n \to \infty} S_n = 5(1-\dfrac{1}{n+1})

=5(1-\dfrac{1}{\infty})\\\\=5(1-0)=5

4 0
3 years ago
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