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Ira Lisetskai [31]
3 years ago
7

65=91/? What is the question mark

Mathematics
1 answer:
trapecia [35]3 years ago
5 0
I would write the missing variable as x. It's what everyone uses.
Anyway, 91 divided by x is 65.
The way we can do this is reverse 65 and x.
91 divided by 65 = x.

91 divided by 1.4 = 65
65 x 1.4 = 91

x (or ?) = 1.4
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Step-by-step explanation:

replace x with 2 and you will got the answer...

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Step-by-step explanation:

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Rewrite the rational exponent as a radical by extending the properties of integer exponents.
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\bf \cfrac{11^{\frac{2}{5}}}{11^{\frac{1}{4}}}\implies 11^{\frac{2}{5}}\cdot 11^{-\frac{1}{4}}\implies 11^{\frac{2}{5}-\frac{1}{4}}\implies 11^{\frac{8-5}{20}}\implies 11^{\frac{3}{20}} \\\\\\ \sqrt[20]{11^3}\implies \sqrt[20]{1331}

5 0
3 years ago
Read 2 more answers
Math6. Grade 6.
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2 (6 x+6)= ?
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8 0
3 years ago
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Use an induction proof to prove this statement:<br> For n≥1, 4^n+5 is divisible by 3.
Tpy6a [65]

Answer:

See below

Step-by-step explanation:

We shall prove that for all n\in\mathbb{N},3|(4^n+5). This tells us that 3 divides 4^n+5 with a remainder of zero.

If we let n=1, then we have 4^{1}+5=9, and evidently, 9|3.

Assume that 4^n+5 is divisible by 3 for n=k, k\in\mathbb{N}. Then, by this assumption, 3|(4^n+5)\Rightarrow4^k+5=3m,\: m\in\mathbb{Z}.

Now, let n=k+1. Then:

4^{k+1}+5=4^k\cdot4+5\\=4^k(3+1)+5\\=3\cdot4^k+4^k+5\\=3\cdot4^k+3m\\=3(4^k+m)

Since 3|(4^k+m), we may conclude, by the axiom of induction, that the property holds for all n\in\mathbb{N}.

3 0
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