Indicate the place value of $2, 6$ and $3$ in $268345$ .

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Solution

Find the place values.
Given number: $268345$
The face value of a number is the value of the digit or number itself. And place value is what the digit represents in its place.
Lakh Ten-thousands Thousand Hundreds Tens Ones
$2$ $6$ $8$ $3$ $4$ $5$
Since, $2$ is in the lakhs digit place.
So, the Place value of $2$ $= 2 \times 100000 = 200000$
Since, $6$ is in the ten-thousands digit place.
So, the Place value of $6$ $= 6 \times 10000 = 60000$
Since, $3$ is in the hundreds digit place.
So, the Place value of $3$ $= 3 \times 100 = 300$
Hence, the place value of $2$ in $268345$ is $200000$ , the place value of $6$ in $268345$ is $60000$ and the place value of $3$ in $268345$ is $300$ .

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Given, $\sqrt{2} = 1.4141 a n d \sqrt{6} = 2.449, f i n d t h e v a l u e o f \frac{1}{\sqrt{3} - \sqrt{2} - 1}$ correct to 3 places of decimal.

2 x - \frac{y}{5} + 6 = 0

\frac{x}{5} - \frac{y}{6} = 1

\sqrt{2} x + \sqrt{3} y = 5

\sqrt{2} = 1.414 and \sqrt{6} = 2.449

\frac{1}{\sqrt{3} - \sqrt{2} - 1}

Express each of the following equations in the form ax + by + c = 0 and indicate the values of a, b, c in each case.

(i) 3x + 5y = 7.5

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(vi) $\sqrt{2} x + \sqrt{3} y = 5$

p (t) = 4 t^{4} + 5 t^{3} - t^{2} + 6

t = a

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