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sergey [27]
3 years ago
8

3 3/5 + 1 1/6 answer as a fraction

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
5 0

Answer:

  • \boxed{\sf{4\dfrac{23}{30} }}

Step-by-step explanation:

It is necessary to add the whole numbers of fractions from left to right in order to solve this problem.

<h3>3 3/5+1 1/6</h3>

<u>First, you have to add the whole numbers.</u>

3+1=4

<u>After that, you can combine by a fraction.</u>

6*3=18

18/5+ 1 1/6

1+6=7

7/6

<u>Rewrite the problem down.</u>

18/5+7/6

<u>Solve.</u>

5*6=30

143/30

=23/30

\boxed{\sf{4\dfrac{23}{30} }}

  • <u>Therefore, the final answer is 4 23/30.</u>
<h3 />

I hope this helps you! Let me know if my answer is wrong or not.

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Step-by-step explanation:

LHS:

\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\ldots \ldots \ldots \ldots \ldots+\frac{1}{\sqrt{8}+\sqrt{9}}1+21+2+31+3+41+……………+8+91

Rationalizing the denominator, we get

\Rightarrow\left(\frac{1}{1+\sqrt{2}} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}+\sqrt{3}} \times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}\right)+\left(\frac{1}{\sqrt{3}+\sqrt{4}} \times \frac{\sqrt{3}-\sqrt{4}}{\sqrt{3}-\sqrt{4}}\right)+\cdots \ldots+\left(\frac{1}{\sqrt{8}+\sqrt{9}} \times \frac{\sqrt{8}-\sqrt{9}}{\sqrt{8}-\sqrt{9}}\right)⇒(1+21×1−21−2)+(2+31×2−32−3)+(3+41×3−43−4)+⋯…+(8+91×8−98−9)

We know that,

\left(a^{2}-b^{2}\right)=(a+b)(a-b)(a2−b2)=(a+b)(a−b)

Now, on substituting the formula, we get,

=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+\cdots \ldots \cdot \frac{(\sqrt{8}-\sqrt{9})}{8-9}=1−21−2+2−32−3+3−43−4+⋯…⋅8−9(8−9)

\Rightarrow \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots+\frac{1}{\sqrt{8}+\sqrt{9}}=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\cdots+(\sqrt{9}-\sqrt{8})⇒1+21+

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