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zhuklara [117]
3 years ago
13

(Multiying by a unit fraction. Find the product mentally. (5.6) × 1/8=

Mathematics
1 answer:
Ipatiy [6.2K]3 years ago
6 0

Step-by-step explanation:

(5.6) × 1/8

(5.6)  \cdot   \frac{1}{8}

Here the unit fraction is 1/8

multiply 5.6 times 1

so its 5.6. then divide the decimal by 8

we know that 8 times 7 is 56

so 56 divide by 8 is 7

we have decimal point 1 number to the left

so we move decimal point 2 place left

Hence 5.6 divide by 8 is 0.7

(5.6)  \cdot   \frac{1}{8}=0.7

Answer: 0.7

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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}

4 0
3 years ago
2) Kari is flying her kite at the park. She has let
Ket [755]

Answer:

approx 58 yard we can determine it by pithagoras theorem

6 0
3 years ago
Si a un múltiplo de 8 le sumo 1 obtengo un múltiplo de 5, ¿qué múltiplos de 8
Simora [160]

Answer:

24, 64

Step-by-step explanation:

If I add 1 to a multiple of 8 I get a multiple of 5, what multiples of 8

meet the condition? (Mention at least 2 of them)

3 x 8 + 1 = 25

8 x 8 + 1 = 65

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3 years ago
What is m∠A?<br> Help please this is overdue
Talja [164]
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6 0
3 years ago
Keisha says that all functions are relations but not all relations are functions. kevin says that all relations are functions bu
vladimir1956 [14]

Keisha is correct, because as per the definition <u>A function is a special relationship where each input has a single output</u>.

A function is a special relation. In other words, a relation if and only if it has a specific characteristic where each input has a single output, then it is called a Function.

All functions are relations but not all relations are functions.


3 0
3 years ago
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