Whats the quotient of 8453 and 35?

Mathematics

1 answer:

klasskru [66]3 years ago

7 0

241.514285715 my friend

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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;

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

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

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The answer to your question is “D”

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Renee is simplifying the expression (7) (StartFraction 13 over 29 EndFraction) (StartFraction 1 over 7 EndFraction). She recogni

dsp73

Answer:

The correct option is commutative property.

Step-by-step explanation:

The expression that Renee is simplifying is:

$(7)\cdot(\frac{13}{29})\cdot(\frac{1}{7})$

It is provided that, Renee recognizes that 7 and $\frac{1}{7}$ are reciprocals, so she would like to find their product before she multiplies by $\frac{13}{29}$ .