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Vera_Pavlovna [14]
3 years ago
11

Help me in number 7 8 9 10 11 12 13 14

Mathematics
2 answers:
Zanzabum3 years ago
8 0
For number 7, you need to find out what number was multiplied by 5 to get 15, which would be 3. So, you multiply the numerator by 3 as well and you will have the missing number.
2/5=6/15

The least common multiple of two numbers is the smallest number that is a multiple of both numbers, that is not zero.
So, you need to find the smallest number that both 14 and 18 can be divided by. 
The answer would be 126.

9 goes into 36 4 times, because 9x4=36. Because of this, you multiply the numerator by 4 as well. 4x5=20.
So, 5/9=20/36

The least common multiple of 5 and 10 would be 10, obviously. It isn't zero, and both 5 and 10 can be divided by 10.

The least common multiple of 42 and 36 is 252.

Three common multiples of 5 and 6. Look at the choices and choose the one where all three numbers can be divided by both 5 and 6.
30/6=5 and 30/5=6, 60/5=12 and 60/6=10, 90/5=18 and 90/6=15.
So, the answer is choice B.

The least common multiple of 30 and 60 is 60. 30 is half of 60, so we already know that 60 can be divided by 30. 60 can be divided by itself as well.

Well, divide each new denominator by 14, and then multiply that number by 9. If the number you get matches the numerator in the choice's fraction, then you found your answer.

Marina CMI [18]3 years ago
3 0
7 is A 8 is A 9 is D 10 is A 11 is
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6.\\\text{We know}\ 9=3^2.\ \text{Therefore}\\\\9^2=(3^2)^2\\\\\text{use}\ (a^n)^m=a^{nm}\\\\9^2=(3^2)^2=3^{2\cdot2}=3^4\\\\\boxed{3^4=9^2}\\=================

7.\\METHOD\ 1:\\\\\sqrt{7^4}=\sqrt{7^{2\cdot2}}\\\\\text{use}\ (a^n)^m=a^{nm}\\\\=\sqrt{(7^2)^2}\\\\\text{use}\ \sqrt{a^2}=a\ \text{for}\ a\geq0\\\\=\boxed{7^2}\\\\METHOD\ 2:\\\\\text{use}\ a^\frac{m}{n}=\sqrt[n]{a^m}\\\\\sqrt{7^4}=7^\frac{4}{2}=\boxed{7^2}\\=================

8.\\5^a\times5^b=5^{11}\\\\a)\\\text{use}\ a^n\times a^m=a^{n+m}\\\\5^a\times5^b=5^{a+b}\\\\5^{a+b}=5^{11}\Rightarrow a+b=11\\\\11\ \text{is odd. The sum of even and odd is odd. Therefore}\ a\ \text{and}\ b\ \text{can't be even.}\\--------------------\\\\b)\\5\ \text{solutions}\\\\11=5+6\\11=4+7\\11=3+8\\11=2+9\\11=1+10

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