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3 years ago

Find the solution to the system of 5x+2y=-24 x-y=5

2 answers:

4 0

So to do this, we have to get rid of one of the variables by adding or subtracting the two equations together.
I will get rid of y. So, we can multiply the second equation by 2. This gives us
2x-2y=10
Now, we can add both equations
This gives us 7x=-14
Therefore, x = -2
Plug that into another equation.
-10 + 2y=-24
Therefore, y = -7

Hope this helped!! :D

Damm [24]3 years ago

4 0

5x + 2y = -24 ⇒ 5x + 2y = -24
1x - 1y = 5 ⇒ 2x - 2y = 10
 7x = -14
 7 7
 x = -2
 x - y = 5
 -2 - y = 5
 + 2 + 2
 -y = 7
 -1 -1
 y = -7
 (x, y) = (-2, -7)

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A quantity P is an exponential function of time I, such that P = 160 when t = 6 and P = 150 when I = 4. Use the given informatio

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Answer:

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Step-by-step explanation:

Perhaps you want to use the points (t, P) = (4, 150) and (6, 160) to find the parameters P0 and k in the equation ...

$P(t)=P_0\cdot e^{kt}$

We know from the given points that we can write the equation as ...

$P(t)=150\left(\dfrac{160}{150}\right)^{(t-4)/(6-4)}=150\left(\dfrac{16}{15}\right)^{\frac{t}{2}-2}\\\\=150\left(\dfrac{16}{15}\right)^{-2}\times\left(\left(\dfrac{16}{15}\right)^{\frac{1}{2}}\right)^t$

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$P_0=150\left(\dfrac{16}{15}\right)^{-2}\approx 131.836\\\\e^{k}=\left(\dfrac{16}{15}\right)^{1/2}\rightarrow k=\dfrac{1}{2}(\ln{16}-\ln{15})\approx 0.0322693$

So, the approximate equation for P is ...

$P(t)=131.836\cdote^{0.032t}$

And the parameters of interest are ...

k = 0.032
P0 = 131.836

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