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3 years ago

What is 5/6 divided by 1/7?

Mathematics

1 answer:

skelet666 [1.2K]3 years ago

8 0

Answer:

5 $\frac{5}{6}$

Step-by-step explanation:

To divide fractions, we multiply the first fraction by the <u>reciprocal </u>of the second fraction. (Seven over one)

$\frac{5*7}{6*1} =\frac{35}{6}$

Now, we convert the fraction into a mixed fraction.

$\frac{35}{6}=5\frac{5}{6}$

<em>The quotient of 5/6 and 1/7 is 5 5/6</em>

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steposvetlana [31]

Answer:

-1 is not correct

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Step-by-step explanation:

Her mistake was adding a negative sign in the one

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3 years ago

The figure below shows a rectangle ABCD having diagonals AC and DB: A rectangle ABCD is shown with diagonals AC and BD. Jimmy wr

yaroslaw [1]

Answer:

C. DC = DC. By reflexive property of equality.

Step-by-step explanation:

We are given that,

In step 4, Jimmy proved ΔADC ≅ ΔBCD by the SAS Congruence Postulate.

For that, the previous steps states,

Step 1: AD = BC, as opposite sides of a rectangle are congruent.

Step 2: ∠ADC = ∠BCD

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Hence, option C is correct.

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3 years ago

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In-s [12.5K]

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Step-by-step explanation:

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2.5x = 5/3

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What is 3hundreds+ 7hunders

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3 years ago

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How do you simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B2%7D%20%2B%5Csqrt%7B6%7D%20%7D%7B%5Csqrt%7B8%7D%20%2B%

blondinia [14]

The trick is to exploit the difference of squares formula,

$a^2-b^2=(a-b)(a+b)$

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

$(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)=(\sqrt8)^2-(\sqrt6)^2=8-6=2$

Whatever you do to the denominator, you have to do to the numerator too. So

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}{(\sqrt8+\sqrt6)(\sqrt8-\sqrt6)}=\dfrac{(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)}2$

Expand the numerator:

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{2\cdot8}+\sqrt{6\cdot8}-\sqrt{2\cdot6}-(\sqrt6)^2$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=\sqrt{16}+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=4+\sqrt{48}-\sqrt{12}-6$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(\sqrt4-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6)=-2+\sqrt{12}(2-1)$

$(\sqrt2+\sqrt6)(\sqrt8-\sqrt6}=-2-\sqrt{12}$

So we have

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+\sqrt{12}}2$

But √12 = √(3•4) = 2√3, so

$\dfrac{\sqrt2+\sqrt6}{\sqrt8+\sqrt6}=-\dfrac{2+2\sqrt3}2=\boxed{-1-\sqrt3}$

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3 years ago