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nikdorinn [45]
3 years ago
7

Must a decimal divisor and a decimal dividend have the same number of decimal places in order to have a whole number quotient? W

rite a division equation using two decimal numbers to support your answer.
Mathematics
1 answer:
fredd [130]3 years ago
8 0
Your answer is B I’m pretty sure about that!
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220 pages in 3 hours
20 pages very 30 minutes so
So 1.5 minutes per page
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3 years ago
Which is the least common multiple of 4 and 12
spayn [35]
The answer is 12. My explanations is that LCM is the smallest number that both numbers can fit into. 12 fits into 12 once, and 4 fits into 12 3 times, so 12 is the answer.
3 0
2 years ago
HELP PLEASE <br><br>i dont understand at all...
bija089 [108]

An exponent signifies repeated multiplication.

x\cdot x\cdot x=x^{3}  the factor x is repeated 3 times

Exponents can be added and subtracted to express the effects of multiplication and division.

\dfrac{x\cdot x\cdot x\cdot x\cdot x}{x\cdot x\cdot x}=\dfrac{x^{5}}{x^{3}}\\\\=\dfrac{x\cdot x\cdot x}{x\cdot x\cdot x}\cdot x\cdot x=x\cdot x\\\\=x^{(5-3)}=x^{2}

The addition and subtraction of exponents works the same even when there are more denominator factors than numerator factors.

\dfrac{x\cdot x\cdot x}{x\cdot x\cdot x\cdot x\cdot x}=\dfrac{x^{3}}{x^{5}}=\dfrac{1}{x^{2}}\\\\=x^{(3-5)}=x^{-2}

That is, a negative numerator exponent is the same as a positive denominator exponent and vice versa. You can move a factor with an exponent from denominator to numerator and change the sign of the exponent, and vice versa.

Your expression has 3 in the denominator with a negative exponent. It can be moved to the numerator and the exponent changed to positive:

\dfrac{1}{3^{-2}}=3^{2}\\\\=3\cdot 3=\bf{9}

8 0
3 years ago
So wait what is the answer then
katrin [286]

Answer:

to what??????

Step-by-step explanation: ???????

4 0
2 years ago
What is the answer to m^7 · m^−3?
elena55 [62]

Answer:

M^4

Step-by-step explanation:

Here use product law which state- same base power should add

M^7-3

M^4

8 0
2 years ago
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