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Vsevolod [243]
4 years ago
5

What are the solutions to the following system of equations? y=x^2+12x+30 and 8x-y=10

Mathematics
1 answer:
Ierofanga [76]4 years ago
6 0
Note that the 2nd equation can be re-written as y=8x-10.
According to the second equation, y=x^2+12x+30.
Equate these two equations to eliminate y:

8x-10 = x^2+12x+30

Group all terms together on the right side.  To do this, add -8x+10 to both sides.  Then 0 = x^2 +4x +40.  You must now solve this quadratic equation for x, if possible.  I found that this equation has NO REAL SOLUTIONS, so we must conclude that the given system of equations has NO REAL SOLUTIONS.

If you have a graphing calculator, please graph 8x-10  and  x^2+12x+30 on the same screen.  You will see two separate graphs that do NOT intersect.  This is another way in which to see / conclude that there is NO REAL SOLUTION to this system of equations.
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Can someone explain why A is the right answer to this question? I'm unsure of how to work out this problem.
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So firstly, we have to find the radius of the circular garden before finding the circumference (the amount of fencing needed to surround the garden). To find the radius, use the area formula (A=\pi r^2), plug in the area of the garden (36 ft^2) and solve for r as such:

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Answer:

the answer is D

Step-by-step explanation:


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