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Sunny_sXe [5.5K]
4 years ago
14

Which arebPartial products for 56x31?

Mathematics
1 answer:
ella [17]4 years ago
4 0
In the problem 56x31, what are the partial products? To acquire the possible partial products we can just multiply the two numbers to produce the possible numbers at hand. <span><span>
1.      </span>56 x 31 = 1736</span><span><span>
2.      </span>31 x 56 = 1736</span><span>   </span>

<span>Same outcome which is explained by the comutative property of multiplication.</span>
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Answer:5

Step-by-step explanation:

Since Michelle (he ? ) has 11/12 quarts of paint, and need 1/6 per quart to paint, we do 11/12 / 1/6 = 11/2, or 5.5. We cannot paint .5 of a block, so Michelle can only paint 5 quarts.

7 0
4 years ago
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Andrew [12]

Answer:

(a)n^{th} = n

f(1) = 1

f(n) = f(n-1) + 1

(b)n^{th} = 2^{n-1}

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(c)n^{th} = n!

f(1) = 1

f(n) = f(n-1) * n

Step-by-step explanation:

(a) This is a sequence of consecutive number

n^{th} = n

f(1) = 1

f(n) = f(n-1) + 1

(b) This is a sequence of 2 to the power of n - 1. The next number is twice time of this number

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f(1) = 1

f(n) = f(n-1) * 2

(c) This is factorial sequence. Where the next number is this number multiplied by n^{th}

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5 0
3 years ago
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Answer:

Values of y are:

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6 0
3 years ago
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evablogger [386]

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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<em>good luck, i hope this helps :)</em>

3 0
3 years ago
Read 2 more answers
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