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Trava [24]
3 years ago
11

PLEASE ANY HELP IT POSSIBLE THANK YOU VERY MUCH ❤

Mathematics
1 answer:
yarga [219]3 years ago
3 0
The hight would be about 3.96

I just put the information given into the formula.

Then divided that by the given volume
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Hannah borrow $30 from her parents. Each week she pays them back the same amount. The total amount she owes her parents after we
natali 33 [55]
30-5d ( where d stands for amount of days, and 5 is the amount of moeay paid per day)
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3 years ago
Write the mixed number as a fraction.<br> 4 1/3
Katyanochek1 [597]
Improper fraction: 13/3
6 0
3 years ago
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Which number should follow in this series: 64/64 9/12 6/12​
zheka24 [161]

Answer: \frac{1}{4}

Step-by-step explanation:

You can reduce the fractions, then:

\frac{64}{64}=1

\frac{9}{12}=\frac{3}{4}

\frac{6}{12}=\frac{3}{6}=\frac{1}{2}

Rewrite them as following:

1,\frac{3}{4},\frac{1}{2}

If you subtract the first number and the second number, you obtain:

1-\frac{3}{4}=\frac{1}{4}

 If you subtract the second number and the second number, you obtain:

\frac{3}{4}-\frac{1}{2}=\frac{1}{4}

Therefore, you must subtract \frac{1}{2} and \frac{1}{4} to obtain the number asked. Then, this is:

\frac{1}{2}-\frac{1}{4}=\frac{1}{4}

6 0
3 years ago
Read 2 more answers
Can 3.65909090909 be expressed as a fraction whose denominator is a power of 10? Explain.
GuDViN [60]
\bf 3.659\textit{ can also be written as }\cfrac{3659}{1000}\textit{ therefore }3.6590909\overline{09}\\\\&#10;\textit{can be written as }\cfrac{3659.0909\overline{09}}{1000}

notice above, all we did, was isolate the "recurring part" to the right of the decimal point, so the repeating 09, ended up on the right of it.

now, let's say, "x" is a variable whose value is the recurring part, therefore then

\bf \cfrac{3659.0909\overline{09}}{1000}\qquad \boxed{x=0.0909\overline{09}} \qquad \cfrac{3659+0.0909\overline{09}}{1000}\implies \cfrac{3659+x}{1000}

now, the idea behind the recurring part is that, we then, once we have it all to the right of the dot, we multiply it by some power of 10, so that it moves it "once" to the left of it, well, the recurring part is 09, is two digits, so let's multiply it by 100 then, 

\bf \begin{array}{llllllll}&#10;100x&=&09.0909\overline{09}\\&#10;&&9+0.0909\overline{09}\\&#10;&&9+x&#10;\end{array}\quad \implies 100x=9+x\implies 99x=9&#10;\\\\\\&#10;x=\cfrac{9}{99}\implies \boxed{x=\cfrac{1}{11}}\\\\&#10;-------------------------------\\\\&#10;\cfrac{3659.0909\overline{09}}{1000}\qquad \boxed{x=0.0909\overline{09}} \quad \cfrac{3659+0.0909\overline{09}}{1000}\implies \cfrac{3659+x}{1000}&#10;\\\\\\&#10;\cfrac{3659+\frac{1}{11}}{1000}

and you can check that in your calculator.
8 0
3 years ago
What is the sixth term in the addition pattern that begins with 15,29,43,57?
WITCHER [35]

Answer:

C is your answer

Step-by-step explanation:

Add every number by 14

5 0
3 years ago
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