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IgorC [24]
3 years ago
10

Please help me with this problem will give branlyiest.

Mathematics
1 answer:
mash [69]3 years ago
7 0

Answer:

$2459.04

$32.31

$33.45 or $34.92

$32.31

$1680.12

$4139.16

$12075.36

20.4% and 3.2%

Step-by-step explanation:

Monthly rent is 204.92, so yearly rent is 204.92 * 12 = 2459.04

His average grocery expenses is the sum of all the grocery expenses, divided by the number of times he did them

33.45 + 34.92 + 28.56 = 96.93

96.93 / 3 = 32.31

33.45 or 34.92

Since he's revising the calculation, and he does groceries once a week, we will multiply the average he spent on grocery by the 52 weeks he's using in calculations. So, 32.31

The estimate for grocery in a year is,

52 * 32.31 = 1680.12

His annual budget for food and rent would be adding the estimated amount he's to spend on grocery together with the estimated amount he's to spend on rent

2459.04 + 1680.12 = 4139.16

If he receives 1006.28 in a month, multiplying by 12 would give us the estimate he's to receive in a year.

1006.28 * 12 = 12075.36

Percentage

Rent = 204.92 / 1006.28 * 100%

Rent = 20.4%

Food = 32.31 / 1006.28 * 100%

Food = 3.2%

Don't forget the brainliest :)

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2 years ago
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. The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to nd
Blababa [14]

The question is:

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution y2(x) of the homogeneous equation

x²y'' - 7xy' + 16y = 0; y1 = x^4

Answer:

The second solution y2 is

A(x^4)lnx

Step-by-step explanation:

Given the homogeneous differential equation

x²y'' - 7xy' + 16y = 0

And a solution: y1 = x^4

We need to find a second solution y2 using the method of reduction of order.

Let y2 = uy1

=> y2 = ux^4

Since y2 is also a solution to the differential equation, it also satisfies it.

Differentiate y2 twice in succession with respect to x, to obtain y2' and y2'' and substitute the resulting values into the original differential equation.

y2' = u'. x^4 + u. 4x³

y2'' = u''. x^4 + u'. 4x³ + u'. 4x³ + u. 12x²

= u''. x^4 + u'. 8x³ + u. 12x²

Now, using these values in the original equation,

x²(u''. x^4 + u'. 8x³ + u. 12x²) - 7x(u'. x^4 + u. 4x³)+ 16(ux^4) = 0

x^6u'' + 8x^5u' + 12x^4u - 7x^5u' - 28x^4u + 16x^4u = 0

x^6u'' + x^5u' = 0

xu'' = -u'

Let w = u'

Then w' = u''

So

xw' = -w

w'/w = -1/x

Integrating both sides

lnw = -lnx + C

w = Ae^(-lnx) (where A = e^C)

w = A/x

But w = u'

So,

u' = A/x

Integrating this

u = Alnx

Since

y2 = uy1

We have

y2 = (Alnx)x^4 = (Ax^4)lnx

Therefore, the second solution y2 is

A(x^4)lnx

6 0
3 years ago
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serious [3.7K]
< BAC = 50
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