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Jobisdone [24]
3 years ago
14

S=

align="absmiddle" class="latex-formula">
(a+l) solve for a
Mathematics
1 answer:
dmitriy555 [2]3 years ago
8 0

\bf S=\cfrac{n}{2}(a+I)\implies S=\cfrac{n(a+I)}{2}\implies 2S=n(a+I)\implies \stackrel{\textit{distributing}}{2S=na+nI} \\\\\\ 2S-nI=na\implies \cfrac{2S-nI}{n}=a

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Which fraction is larger 2/3 or 15/18
kirza4 [7]
The answer is 15/18
because if you just put 15/18 it will give you <span>0.83333333333
and for 2/3 it will give you </span><span>0.66666666666
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7 0
2 years ago
Which of the following is a composite number?<br> A. 19<br> B. 63<br> C. 1<br> D. 0
Vikentia [17]
Which of the following is a composite number?<span>A. 19
B. 63
C. 1
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6 0
3 years ago
For the standard form of two billion three Hundred fifty thousand four
Drupady [299]
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5 0
3 years ago
Solve for x: one over five (5x + 12) = 18 (2 points)
Usimov [2.4K]
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8 0
2 years ago
Which of the following statements are true? Select all that apply.
Tom [10]

Answer: The statements in options 2 and 4 are true.

Explanation:

The LCM of two numbers is the smallest number that is multiple of both numbers.

The factor of 8 and 12 are,

8=2\times 2\times 2

12=2\times 2\times 3

Since 2 and 2 are common in both but 2 and 3 are not same, therefore the L.C.M. of 8 and 12 is,

L.C.M.(8,12)=2\times 2\times 2\times 3=24

Therefore the statement in option 1 is incorrect.

The factor of 6 and 9 are,

6=2\times 3

9=3\times 3

Since 3 is common in both but 2 and 3 are not same, therefore the L.C.M. of 6 and 9 is,

L.C.M.(6,9)=2\times 3\times 3=18

Therefore the statement in option 2 is correct.

The factor of 11 and 4 are,

11=1\times 11

4=2\times 2

Since all factor are different, therefore the L.C.M. of 11 and 4 is,

L.C.M.(11,4)=1\times 11\times 2\times 2=44

Therefore the statement in option 3 is incorrect.

The factor of 9 and 10 are,

9=1\times 9

10=2\times 5

Since all factor are different, therefore the L.C.M. of 9 and 10 is,

L.C.M.(9,10)=1\times 9\times 2\times 5=90

Therefore the statement in option 4 is correct.

4 0
3 years ago
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