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elixir [45]
4 years ago
10

PLEASE HELP!

Mathematics
1 answer:
Roman55 [17]4 years ago
6 0
There are various ways to solve this. I will solve by elimination which seems easier:
I multiply the top equation by 1/2 to get 1/4y so I can eliminate y:
-3x/8 + 1y/4 = -1 Now add the second equation:
1x/5 - 1y/4 = -6 This gives:
------------------------------------
-3x/8 + 1x/5 = -7 Get a common denominator:
-15x/40 + 8x/40 = -7 Now add fractions:
-7x/40 = -7 Multiply by the reciprocal(-40/7):
x = 7 × 40 / 7 Simplify:
x = 40.
Now plug in 40 for x in one of the equations and solve for y (2nd equation looks easier):
1(40)/5 - 1y/4 = -6
8 - y/4 = -6 subtract 8
-y/4 = -14 multiply by 4
-y = -56
y = 56.

So we got x=40 and y=(56) new lets check!
-3(40)/4 + (56)/2 = -2
-3(10) + 28 = -2
-30 + 28 = -2
-2 = -2.
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evablogger [386]
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.. a rectangle 2 mi x 5 mi at upper left.
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Then the sum of areas is
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4 0
3 years ago
(15 pts) 4. Find the solution of the following initial value problem: y"-10y'+25y = 0 with y(0) = 3 and y'(0) = 13
jolli1 [7]

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Step-by-step explanation:

The given differential equation is y''-10y'+25y=0

The characteristics equation is given by

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Finding the values of r

r^2-5r-5r+25=0\\\\r(r-5)-5(r-5)=0\\\\(r-5)(r-5)=0\\\\r_{1,2}=5

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y'(x)=5c_1e^{5x}+5c_2xe^{5x}+c_2e^{5x}...(ii)

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Now, apply the initial condition y' (0)= 13 in equation (ii)

13=5(3)e^{0}+0+c_2e^{0}\\\\13=15+c_2\\\\c_2=-2

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3 years ago
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Vsevolod [243]

Answer:

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Step-by-step explanation:

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3 years ago
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