What is 1/100 of 650

Mathematics

1 answer:

marishachu [46]4 years ago

5 0

This is easy young one it is 6.5 because you move the decimal place 2 places to the left when dividing by 100
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7. By using binomial expansion show that the value of (1.01)^12 exceed the value of (1.02)^6 by 0.0007 correct to four decimal p

BlackZzzverrR [31]

Binomial expansion is used to factor expressions that can be expressed as the power of the sum of two numbers.

The proof that (1.01)^12 exceeds (1.02)^6 by 0.0007 is $\mathbf{(1.01)^{12} - (1.02)^6 \approx 0.0007 }$

The expressions are given as:

$\mathbf{(1.01)^{12}\ and\ (1.02)^6}$

A binomial expression is represented as:

$\mathbf{(a + b)^n = \sum\limits^n_{k=0}^nC_k a^{n - k}b^k}$

Express 1.01 as 1 + 0.01

So, we have:

$\mathbf{(1.01)^{12} = (1 + 0.01)^{12}}$

Apply the above formula

$\mathbf{(1.01)^{12} = ^{12}C_0 \times 1^{12 - 0} \times 0.01^0 + ......... .......... + ^{12}C_{12} \times 1^{12 - 12} \times 0.01^{12} }}$

$\mathbf{(1.01)^{12} = 1 \times 1 \times 1 + ......... .......... + 1 \times 1 \times 10^{-24} }}$

$\mathbf{(1.01)^{12} = 1 + ......... .......... + 10^{-24} }}$

This gives

$\mathbf{(1.01)^{12} = 1.1268\ (approximated)}$

Similarly,

Express 1.02 as 1 + 0.02

So, we have:

$\mathbf{(1.02)^6 = (1 + 0.02)^6}$

Apply $\mathbf{(a + b)^n = \sum\limits^n_{k=0}^nC_k a^{n - k}b^k}$

$\mathbf{(1.02)^6 = ^6C_0 \times 1^{6 - 0} \times 0.02^0 + ^6C_1 \times 1^{6 - 1} \times 0.02^1 +.............. + ^6C_6 \times 1^{6 - 6} \times 0.02^6 }$ $\mathbf{(1.02)^6 = 1 \times 1 \times 1 + 6 \times 1 \times 0.02 +.............. + 1 \times 1 \times 6.4 \times 10^{-11} }$