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3 years ago

11

How many times larger is 5.4x10^15 than 9.0x10^3?

1 answer:

KengaRu [80]3 years ago

4 0

Answer:

6 * 10^11

Step-by-step explanation:

how many times larger is 5.4x10^15 than 9.0x10^3?

So, we need to divide 5.4x10^15 over 9.0x10^3

So, let it = x

$x= \frac{5.4*10^{15}}{9*10^3} = \frac{54*10^{15}}{10*9*10^3} = \frac{54*10^{15}}{9*10^4}=\frac{54}{9}*10^{15-4}=6*10^{11}$

Note: 5.4 = 54/10

$a^m * a^n = a^{m+n}\\a^m / a^n = a^{m-n}$

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Answer:

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3 years ago

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3 years ago

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kondor19780726 [428]

Answer:

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Step-by-step explanation:

Hi!

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3 years ago

Please answer the question in the photo, thanks!

belka [17]

Answer:

$\therefore 2\dfrac{1}{3} \div \left ( -3\dfrac{2}{3} \right ) = -\dfrac{7}{11}$

Step-by-step explanation:

The question relates to dividing a fraction by proper and another fraction

The given expression is presented as follows;

$2\dfrac{1}{3} \div \left ( -3\dfrac{2}{3} \right )$

We rearrange the fractions as improper fractions for easier division as follows;

$2\dfrac{1}{3} = \dfrac{7}{3}$

$-3\dfrac{2}{3} = -\dfrac{11}{3}$

Therefore, we have;

$2\dfrac{1}{3} \div \left ( -3\dfrac{2}{3} \right ) = \dfrac{7}{3} \div \left (-\dfrac{11}{3} \right ) = \dfrac{7}{3} \times \left (\dfrac{1}{-\dfrac{11}{3} } \right ) = \dfrac{7}{3} \div \left (-\dfrac{3}{11} \right ) =-\dfrac{7}{11}$

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3 years ago

Fill in the blanks with two consecutive integers to complete the following inequality.

Alona [7]

Sqrt of 56 is about 7.5

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3 years ago

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