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stira [4]
3 years ago
7

The table lists how financial aid income cutoffs (in dollars) for a family of four have changed over time. Use the midpoint form

ula to approximate the financial aid cutoff for 1985.
Year. Income
1960. 21,250
1970. 27,500
1980. 33,750
1990. 40,000
2000. 46,250
Mathematics
2 answers:
crimeas [40]3 years ago
4 0

Answer:

is go listen to high fashion

Step-by-step explanation:

5t6 67nn jfuuk

Kipish [7]3 years ago
3 0
You can see that every decade the income increases by 6,250. So if you want to do 5 years instead you would have to add 3,125 to the income of 1980 to find what the income was for 1985. Now you add 33,750+3,125= 36,875. The financial aid cutoff for 1985 would be $36,875.
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Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x
djverab [1.8K]

I'll assume the ODE is

y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}

Solve the homogeneous ODE,

y'' - 3y' + 2y = 0

The characteristic equation

r^2 - 3r + 2 = (r - 1) (r - 2) = 0

has roots at r=1 and r=2. Then the characteristic solution is

y = C_1 e^x + C_2 e^{2x}

For nonhomogeneous ODE (1),

y'' - 3y' + 2y = e^x

consider the ansatz particular solution

y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x

Substituting this into (1) gives

a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1

For the nonhomogeneous ODE (2),

y'' - 3y' + 2y = e^{2x}

take the ansatz

y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}

Substitute (2) into the ODE to get

b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1

Lastly, for the nonhomogeneous ODE (3)

y'' - 3y' + 2y = e^{-x}

take the ansatz

y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}

and solve for c.

ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16

Then the general solution to the ODE is

\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}

6 0
1 year ago
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Svetach [21]
195 sq meters.  Area of triangle is......

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4 0
3 years ago
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Lelu [443]

Answer:

Given that JN was bisected, JL ≅ LN

Given that KM was bisected, KL ≅ ML

∠JLK ≅ ∠MLN because of vertical angles.

∠JLK is contained by JL and KL.

∠MLN is contained by ML and LN.

Therefore ΔJKL ≅ ΔNML by the SAS postulate.

Step-by-step explanation:

The SAS postulate states that when you know two triangles have an equal angle, and that angle is formed by two sides that are equal in both triangles, the two triangles are congruent.

When a line is bisected, it means it was cut in two equal parts.

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7 0
3 years ago
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Rudik [331]

let the consecutive multiples be 7(n-1) , 7n and 7(n+1)

so 7(n-1)+7n+7(n+1)=777

or 3n=111,

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3 years ago
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Firdavs [7]

Answer:

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13:30

Ratio of girls to all students:

17:30

Ratio of boys to girls:

13:17

Step-by-step explanation:

I think you're asking for the ratio... so I hope this helps! :D

7 0
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