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Hunter-Best [27]
3 years ago
14

Find the GCF of each pair of numbers.

Mathematics
1 answer:
Artist 52 [7]3 years ago
7 0
<h3>Find the GCF of each pair of numbers.</h3>

Answer 9: 40 and 48 = (2³ x 5) and (2⁴ x 3) = 2³ = 8✔️

Answer 10: 30 and 45 = (2 x 3 x 5) and (3² x 5) = 3 x 5 = 15✔️

Answer 11: 10 and 45 = (2 x 5) and (3² x 5) = 5✔️

Answer 12: 25 and 90 = (5 x 5) and (2² x 3 x 5) = 5✔️

Answer 14: 28 and 70 = (2² x 7) and (2 x 5 x 7) = 2 x 7 = 14✔️

Answer 13: 21 and 40 = (3 x 7) and (2³ x 5) = 1✔️

Answer 15: 60 and 72 = (2² x 3 x 5) and (2³ x 3²) = 2² x 3 = 12✔️

Answer 16: 45 and 81 = (3² x 5) and (3⁴) = 3² = 9✔️

Answer 18: 55 and 77 = (5 x 11) and (7 x 11) = 11✔️

Answer 17: 28 and 32 = (2² x 7) and (2⁵) = 2² = 4✔️

Step-by-step explanation: First, find the prime factorizations of the two numbers. Then, find the common factors raised to the smallest exponent and multiply each other. If there are no common prime factors, the GCF is 1.

\textit{\textbf{Spymore}}

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<h3>s³=l,b,h</h3><h3>so,</h3><h3>s³=125</h3>

s =  \sqrt[3]{125}

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Using two normal dice, what is the probability of rolling numbers that add up to 3?
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so possibilities are 1 on first die and 2 on the second die

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Bas_tet [7]

E-3 will be the first student to start writing the assignment.

It is clearly stated in the statement that the order of starting the assignment and the order of the completion of the assignment is not the same, the digit in the name of the executive is also not same. This information puts some restrictions.

E-1 can neither be the first one to start nor the first one to complete. Similarly, E-2 cannot be the second one and E-3 cannot be the third one.

It is given that the last student (the third one) to start is the first student (the first one) to complete. This can neither be E-3 nor E-1. It is E-2

The final arrangement is as follows.

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First                           E-3              E-2

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E-3 is the first one to start writing the assignment.

More similar problems are solved in the link.

brainly.com/question/28391619?referrer=searchResults

#SPJ4

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