Subtract the following:
$4 2 2 . 3 2 + 1 2 1 . 8 1$

Open in App

Solution

Subtraction of decimals
The given decimals are already into like decimals.
So, arrange them according to place value,
$H T O t h 4 2 2 . 3 2 + 1 2 1 . 8 1$
Therefore,
$H T O t h 1 13 4 2 2 . 3 2 + 1 2 1 . 8 1 3 0 0 . 5 1$

Suggest Corrections

Similar questions

Q. Find the values of the following:

(1) 11³
(2) (–5)²
(3) (–3)³
(4) 10⁷
(5) 2⁴
(6)

{(\frac{2}{3})}^{5}

(7)

{(- \frac{1}{5})}^{4}

(8) 2⁸
(9) (–1)⁸
(10) (–1)⁹

Find x, if

$(i) {(\frac{1}{4})}^{- 4} \times {(\frac{1}{4})}^{- 8} = {(\frac{1}{4})}^{- 4 x} (i i) {(\frac{- 1}{2})}^{- 19} \div {(\frac{- 1}{2})}^{8} = {(\frac{- 1}{2})}^{- 2 x + 1} (i i i) {(\frac{3}{2})}^{- 3} \div {(\frac{3}{2})}^{5} = {(\frac{3}{2})}^{2 x + 1} (i v) {(\frac{2}{5})}^{- 3} \times {(\frac{2}{5})}^{15} = {(\frac{2}{5})}^{2 + 3 x} (v) {(\frac{5}{4})}^{- x} \div {(\frac{5}{4})}^{- 4} = {(\frac{5}{4})}^{5} (v i) {(\frac{8}{3})}^{2 x + 1} \times {(\frac{8}{3})}^{5} = {(\frac{8}{3})}^{x + 2}$

Subtract the following:

$8 . 2 1 - 4 . 6 7$

Subtract the following:

$4 4 . 3 3 - 1 2 . 7 6$

Q. Which of the followingis a factor of (2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1) options(1)2^4-4 (2)2^4-1 (3)2^8+8 (4)2^2+2

Join BYJU'S Learning Program

Subtract the following:422.32+121.81

Subtraction of decimalsThe given decimals are already into like decimals.So, arrange them according to place value,HTOth422.32+121.81Therefore, HTOth113422.32+121.81300.51

Subtract the following:
$4 2 2 . 3 2 + 1 2 1 . 8 1$

Subtraction of decimals
The given decimals are already into like decimals.
So, arrange them according to place value,
$H T O t h 4 2 2 . 3 2 + 1 2 1 . 8 1$
Therefore,
$H T O t h 1 13 4 2 2 . 3 2 + 1 2 1 . 8 1 3 0 0 . 5 1$