Answer:
B
Step-by-step explanation:
Two negatives is a positive, and you want your outcome to be a positive x, therefore b is the correct answer.
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I'd suggest using "elimination by addition and subtraction" here, altho' there are other approaches (such as matrices, substitution, etc.).
Note that if you add the 3rd equation to the second, the x terms cancel out, and you are left with the system
- y + 3z = -2
y + z = -2
-----------------
4z = -4, so z = -1.
Next, multiply the 3rd equation by 2: You'll get -2x + 2y + 2z = -2.
Add this result to the first equation. The 2x terms will cancel, leaving you with the system
2y + 2z = -2
y + z = 4
This would be a good time to subst. -1 for z. We then get:
-2y - 2 = -2. Then y must be 0. y = 0.
Now subst. -1 for z and 0 for y in any of the original equations.
For example, x - (-1) + 3(0) = -2, so x + 1 = -2, or x = -3.
Then a tentative solution is (-3, -1, 0).
It's very important that you ensure that this satisfies all 3 of the originale quations.
graph B shows the correct business equation.
we can represent it as
y = 5 + 2x
as x = 1 , y = 5+2×1=5+2=7
and so on.
at the beginning of year 4, only 3 years have elapsed, the 4th year hasn't started yet, since it's at the beginning, so at the beginning of year 4 we can say only 4-1 years have elapsed.
![A=700\left( 1 + \frac{0.05}{1} \right)^{1\cdot 3}\implies A = 700(1+0.05)^3\implies A(4)=700(1+0.05)^{4-1} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill A(n)=700(1+0.05)^{n-1}~\hfill](https://tex.z-dn.net/?f=A%3D700%5Cleft%28%201%20%2B%20%5Cfrac%7B0.05%7D%7B1%7D%20%5Cright%29%5E%7B1%5Ccdot%203%7D%5Cimplies%20A%20%3D%20700%281%2B0.05%29%5E3%5Cimplies%20A%284%29%3D700%281%2B0.05%29%5E%7B4-1%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20A%28n%29%3D700%281%2B0.05%29%5E%7Bn-1%7D~%5Chfill)
