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DerKrebs [107]
2 years ago
6

PLEASE HELP ME: the ratio of cars to trucks on a highway at any given time is 12 to 8. If there are always more than 105 cars on

the highway, what is the minimum number of cars that can be on the highway?
Mathematics
1 answer:
andre [41]2 years ago
5 0

Answer:

70

Step-by-step explanation:

First, set up the ratio of c:t, which is 12:8. Then set up your new ratio of c:t, which is 105:?. The way I find the easiest is to divide 105 by 12, which is 8.75, then multiply that by 8, which is 70. To check if it's right, just divide 12 by 8, and 105 by 70, and if they're the same number, you've got it right.

Ps. I'm guessing you meant to ask the minimum number of trucks, not cars.

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olasank [31]

Answer:

y=\frac{-6){5}x-3

Step-by-step explanation:

given 6x+5y=-15

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y=\frac{-6}{5}x-\frac{15}{5}

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slope=[tex]\frac{-6}{5}[tex], intercept=-3

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Last quarter, 84 seventh grade students made honor rolls. if there are 400 students in the seventh grade, what percent made hono
lisov135 [29]
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7 0
2 years ago
(D^2 +2D +1)y=e^-x log(x) solve using method variation of parameter?
Ad libitum [116K]
First you'll need the complementary solutions. The characteristic equation for this ODE is

D^2+2D+1=(D+1)^2=0\implies D=-1

with multiplicity 2. This means two linearly independent solutions will be y_1=e^{-x} and y_2=xe^{-x}.

Via variation of parameters, the particular solution will take the form

y_p=y_1u_1+y_2u_2

where

u_1=-\displaystyle\int\frac{y_2e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1e^{-x}\log x}{W(y_1,y_2)}\,\mathrm dx

The Wronskian is

W(y_1,y_2)=\begin{vmatrix}e^{-x}&xe^{-x}\\-e^{-x}&e^{-x}(1-x)\end{vmatrix}=e^{-2x}(1-x)+xe^{-2x}=e^{-2x}

So, you have

u_1=-\displaystyle\int\frac{xe^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx
u_1=-\displaystyle\int x\log x\,\mathrm dx
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and

u_2=\displaystyle\int\frac{e^{-x}e^{-x}\log x}{e^{-2x}}\,\mathrm dx
u_2=\displaystyle\int\log x\,\mathrm dx
u_2=x(\log x-1)

So the solution to the ODE is

y=C_1y_1+C_2y_2+y_1u_1+y_2u_2
y=(C_1+C_2x)e^{-x}+\left(\dfrac14x^2-\dfrac12x^2\log x\right)e^{-x}+(\log x-1)x^2e^{-x}
y=(C_1+C_2x)e^{-x}+\dfrac14x^2e^{-x}(2\log x-3)
3 0
3 years ago
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