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3 years ago

HELP NEEDED ASAP!!!!

Mathematics

1 answer:

LenKa [72]3 years ago

3 0

I don’t know the answer for a and b. I’m really sorry.

C) 24 more

D) 18 more hours

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3 years ago

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Roman55 [17]

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3 years ago

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A shirt that normally costs $30 is on sale for $21.75 what percent of the regular price is the sale price

Phoenix [80]

First Divide: 21.75/ 30

You will get 0.725

Then, multiply 0.725 by 100%

Then your answer should be: 72.5%

3 0

3 years ago

The lines in the graph of part A don't intersect at two whole numbers on the coordinate grid. Estimate the values of the solutio

mariarad [96]

Answer:

The solution lies at approximately (4.25,6.5).

Step-by-step explanation:

The estimate is pretty close for both equations. For the first equation, the estimate results in a value that is only 0.2125 units away from the constant. For the second equation, the estimate results in a value that is only 0.25 units away from the constant.

5 0

3 years ago

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.

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

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

hence the complete solution

y = C.F. + P.I.

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

7 0

3 years ago