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LuckyWell [14K]

2 years ago

9

A coat that originally sold for $369.00 is on sale at a 40% discount? How much will you pay for the coat?

1 answer:

zaharov [31]2 years ago

3 0

We calculate 40% from 369,00 $, this is discount

40%*369= $\frac{40}{100}*369= 147,60$

After sale coat costs 147,60 less so:

369 -147,60=221,40 $

Now I will pay for the coat 221,40$.

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2) Rachel wanted to drink exactly three bottles of water each day, so she bought twenty-six

lesya [120]

Answer:

1

Step-by-step explanation:

Twenty-Six is not a multiple of three so you know it won't be 26. Counting by threes goes 3, 6, 9, 12, 15, 18, 21, 24, 27, 30. Twenty-seven is the closest number to 26 which is higher than 26. Twenty-six + 1 = 27. That means that 1 is the answer.

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janice correctly answered 21 of the 24 questions. what percent of the questions did she answer correctly?

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She got 87.5 percent. you find the answer by setting up a proportion

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2 years ago

The eighth grade class is planning a trip to Washington DC this spring. They need a minimum of 75 people to attend in order to r

8_murik_8 [283]

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2 years ago

Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

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2 years ago

zhuklara [117]

Answer: -30

Step-by-step explanation: -13 subtract 17 = -30

5 0

2 years ago