12 divided by 2 x the absolute value of 1.3 minus 2.16

Mathematics

1 answer:

yaroslaw [1]1 year ago

7 0

The value of the expression (12/2)×| 1.3 - 2.16 | is 5.16 .

The Absolute Value of a number means writing the number without the negative sign .

For Example : the absolute value of -2 is |-2| = 2

In the question ,

an expression in words is given as "12 divided by 2 x the absolute value of 1.3 minus 2.16 "

the above expression in numbers can be represented by

= (12/2)×| 1.3 - 2.16 |

= 6 × | 1.3 - 2.16 |

= 6 × |-0.86|

= 6 × 0.86 .... because |-0.86| = 0.86

= 5.16

Therefore , The value of the expression (12/2)×| 1.3 - 2.16 | is 5.16 .

The given question is incomplete , the complete question is

Find the value of the expression " 12 divided by 2 x the absolute value of 1.3 minus 2.16 " ?

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Solve the following ODE's: c) y* - 9y' + 18y = t^2

Nastasia [14]

Answer:

y = $C_1e^{3t}+C_2e^{6t}$ + $\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$

Step-by-step explanation:

y''- 9 y' + 18 y = t²

solution of ordinary differential equation

using characteristics equation

m² - 9 m + 18 = 0

m² - 3 m - 6 m+ 18 = 0

(m-3)(m-6) = 0

m = 3,6

C.F. = $C_1e^{3t}+C_2e^{6t}$

now calculating P.I.

$P.I. = \frac{t^2}{D^2 - 9D +18}$

$P.I. = \dfrac{t^2}{(D-3)(D-6)}\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1-\frac{D}{6})^{-1}(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(1+\frac{D}{6}+\frac{D^2}{36}+....)(t^2)\\P.I. =\dfrac{1}{18}(1-\frac{D}{3})^{-1}(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(1+\frac{D}{3}+\frac{D^2}{9}+....)(t^2+\frac{2t}{6} + \frac{2}{36})\\P.I. =\dfrac{1}{18}(t^2+\frac{2t}{6} + \frac{2}{36}+\frac{2t}{3}+\frac{2}{18}+\frac{2}{9})$