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larisa86 [58]
3 years ago
5

Solve the following system of equation graphically 2x+y=-1 and x+2y=4

Mathematics
1 answer:
Mumz [18]3 years ago
4 0
As 2x + y = -1

Then,

y = - 1 - 2x

We can to replace on the bellow equation.

x + 2y = 4

x + 2.( -1 -2x) = 4

Doing the distributive

x + 2.(-1) + 2.(-2x) = 4

x -2 -4x = 4

-3x -2 = 4

-3x = 4 + 2

-3x = 6

x = 6/-3

x = -2
________________

Then, the value of y will be:

y = -1 -2x

y = -1 -2.(-2)

y = -1 + 4

y = 3
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\frac{19}{6} many more miles of road are left to build .

<u>Step-by-step explanation:</u>

Here we have , A construction crew must build 4 miles of road in one week. On Monday, they build 1/2 mile of road. On Tuesday, they build 1/3 mile of road.  We need to find that  How many more miles of road are left to build . Let's find out:

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Let Miles of road left to build is x So,

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3 years ago
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The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s
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We are given the following function:

s(t)=2t^{3}-21t^{2}+60t+3

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v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s

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Now, substituting the times to calculate the accelerations, we have:

a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}

Now, substitutitng the times to calculate the positions, we have:

s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft

Have a nice day!

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