Question 1v7-10-2-5-3-2

Name: NCERT - Grade 07 - Mathematics - Fractions and Decimals - Q01_5
Uploaded: 2022-07-04T10:36:42+05:30
Description: 7/10 + 2/5 + 3/2 = 7 + 4 + 15/10 = 26/10 = 13/5 = 23/5 ...

7/10 + 2/5 + 3/2 = 7 + 4 + 15/10 = 26/10 = 13/5 = 23/5 ...

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Byju's Answer

Standard IX

Mathematics

Complement of a Set

Question 1v7/...

Question

Question 1 (v)
$\frac{7}{10} + \frac{2}{5} + \frac{3}{2}$

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Solution

$\frac{7}{10} + \frac{2}{5} + \frac{3}{2} = \frac{7 + 4 + 15}{10} = \frac{26}{10} = \frac{13}{5} = 2 \frac{3}{5}$

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Q. Question 2
Simplify:

\frac{7 \sqrt{3}}{\sqrt{10} + \sqrt{3}} - \frac{2 \sqrt{5}}{\sqrt{6} + \sqrt{5}} - \frac{3 \sqrt{2}}{\sqrt{15} + 3 \sqrt{2}}

Q. Question 2
Simplify:

\frac{7 \sqrt{3}}{\sqrt{10} + \sqrt{3}} - \frac{2 \sqrt{5}}{\sqrt{6} + \sqrt{5}} - \frac{3 \sqrt{2}}{\sqrt{15} + 3 \sqrt{2}}

Question 43

2__ $\frac{5}{3} = \frac{10}{3}$

Question 3 (vi)

Subtract:

$- x^{2} + 10 x - 5 f r o m 5 x - 10$

Q. Question 2(b)
Find the number from each of the following expanded forms:
(b)

4 \times 10^{5} + 5 \times 10^{3} + 3 \times 10^{2} + 2 \times 10^{0}

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Complement of a Set

Standard IX Mathematics

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Question 1 (v) 710+25+32

710+25+32=7+4+1510=2610=135=235

Question 1 (v)
$\frac{7}{10} + \frac{2}{5} + \frac{3}{2}$

$\frac{7}{10} + \frac{2}{5} + \frac{3}{2} = \frac{7 + 4 + 15}{10} = \frac{26}{10} = \frac{13}{5} = 2 \frac{3}{5}$