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Andreas93 [3]
3 years ago
10

Denise sold 24 more dresses in January than in February and 10 fewer dresses in March than in January. In March she sold 28 dres

ses. How many dresses did Denise sell in February?
Mathematics
1 answer:
Svet_ta [14]3 years ago
8 0
Denise sold 14 dresses in February.
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I need some help please What's |5| + |–7|?
luda_lava [24]

Answer:

<h3>hello there!</h3>

\left|5\right|+\left|-7\right|

rule =a,\:a\ge 0

|5|=5

\mathrm{Apply\:absolute\:rule}:\quad \left|-a\right|=a

\left|-7\right|=7

5+7=

12

hope this helps you..

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Given that the following statement is true
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A housekeeper had 4/5 of a spray bottle of a cleaning solution. He used 1/8 of the solution in a day.How much of the bottle did
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How do you do this question?
ElenaW [278]

x*y' + y = 8x

y' + y/x = 8 .... divide everything by x

dy/dx + y/x = 8

dy/dx + (1/x)*y = 8

We have something in the form

y' + P(x)*y = Q(x)

which is a first order ODE

The integrating factor is u(x) = e^{\int P(x)dx} = e^{\int (1/x) dx} = e^{\ln(x)} = x

Multiply both sides by the integrating factor (x) and we get the following:

dy/dx + (1/x)*y = 8

x*dy/dx + x*(1/x)*y = x*8

x*dy/dx + y = 8x

y + x*dy/dx = 8x

Note the left hand side is the result of using the product rule on xy. We technically didn't need the integrating factor since we already had the original equation in this format, but I wanted to use it anyway (since other ODE problems may not be as simple).

Since (xy)' turns into y + x*dy/dx, and vice versa, this means

y + x*dy/dx = 8x turns into (xy)' = 8x

Integrating both sides with respect to x leads to

xy = 4x^2 + C

y = (4x^2 + C)/x

y = (4x^2)/x + C/x

y = 4x + Cx^(-1)

where C is a constant. In this case, C = -5 leads to a solution

y = 4x - 5x^(-1)

you can check this answer by deriving both sides with respect to x

dy/dx = 4 + 5x^(-2)

Then plugging this along with y = 4x - 5x^(-1) into the ODE given, and you should find it satisfies that equation.

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3 years ago
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