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4 years ago

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90=20+70 use the distributive property and the GCF OF 20 and 70 to write another related expression for 90 could you write anoth

er expression with a different common factor?

2 answers:

ziro4ka [17]4 years ago

7 0

90= 10 (2+7)
10 is the common factor of 70 and 20

mixas84 [53]4 years ago

6 0

Answer:

$90=10(2+7)$

Yes, we can write another expression for 90 with a different common factor

$90=5(4+14)$

Step-by-step explanation:

We are given that

90=20+70

We have to use distributive property amd GCF Of 20 and 70 .

We have to write an expression for 90 and another expression with different common factor.

$20=2\times 2\times 5$

$70=2\times 5\times 7$

GCF (20,70)= $2\times 5=10$

$90=10(2+7)$

Now, another expression

$90=5(4+14)$

Yes, we could write another expression with a different common factor.

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Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

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$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

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Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$

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In concusion, the statement is true if and only if n is a non negative integer such that $n\leq 77$

So, 78 is the smallest value of n that does not satisfy the inequality.

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$(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064$

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Reason

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