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3 years ago

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Sarah's mother gives her $25 for each A she gets on her report card and $10 for each B. If Sarah's mother gave her $80 and Sarah

had 3 B's, how many A's did she have on her report card?

1 answer:

iragen [17]3 years ago

4 0

Answer:

2 A"s

Step-by-step explanation:

You take 80-30 and are left with 50 so you divide that by 25 and get 2

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Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).

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Answer:

A) Particular solution:

$2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

B) Homogeneous solution:

$y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))$

C) The most general solution is

$y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

Step-by-step explanation:

Given non homogeneous ODE is

$y''+4y'+5y=10x+e^{-x}---(1)$

To find homogeneous solution:

$D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)$

To find particular solution:

$y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\$

Substituting $y_{p},y'_{p},y''_{p}$ in (1)

$y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\$

Equating the coefficients

$5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\$

The general solution is

$y=y_{h}+y_{p}$

from (2) ad (3)

$y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}$

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Please help me on this it’s due

olya-2409 [2.1K]

<h3>Answer: 5 cakes</h3>

================================================

Explanation:

Let's start off converting the mixed number 12 & 1/4 to an improper fraction.

$a \frac{b}{c} = \frac{a*c+b}{c}\\\\12 \frac{1}{4} = \frac{12*4+1}{4}\\\\12 \frac{1}{4} = \frac{49}{4}\\\\$

Do the same for the other mixed number 2 & 1/3.

$a \frac{b}{c} = \frac{a*c+b}{c}\\\\2 \frac{1}{3} = \frac{2*3+1}{3}\\\\2 \frac{1}{3} = \frac{7}{3}\\\\$

-----------------------

From here, we divide the two fractions. I converted them to improper fractions to make the division process easier.

$\frac{49}{4} \div \frac{7}{3} = \frac{49}{4} \times \frac{3}{7}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{49\times 3}{4\times 7}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{7\times 7\times 3}{4\times 7}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{7\times 3}{4}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{21}{4}\\\\$

The last step is to convert that result to a mixed number.

$\frac{21}{4} = \frac{4*5+1}{4}\\\\\frac{21}{4} = \frac{4*5}{4}+\frac{1}{4}\\\\\frac{21}{4} = 4+\frac{1}{4}\\\\\frac{21}{4} = 5 \frac{1}{4}\\\\$

Note that 21/4 = 5.25 and 1/4 = 0.25 to help check the answer.

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Therefore, she can make 5 cakes. The fractional portion 1/4 is ignored since we're only considering whole cakes rather than partial ones.

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