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Artemon [7]
3 years ago
6

Solve system by elimination y=x^2 y=x+2 SHOW YOUR WORK

Mathematics
1 answer:
Dimas [21]3 years ago
3 0
We have that
y=x²----> equation 1<span>
y=x+2-----> equation 2

multiply equation 1 by -1
-y=-x</span>²

add equation 1 and equation 2
-y=-x²
 y=x+2
------------
0=-x²+x+2-------------> -x²+x+2=0-----> x²-x-2=0

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(x²-x)=2
<span>Complete the square. Remember to balance the equation by adding the same constants to each side
</span>(x²-x+0.5²)=2+0.5²

Rewrite as perfect squares

(x-0.5)²=2+0.5²
(x-0.5)²=2.25-----> (x-0.5)=(+/-)√2.25-----> (x-0.5)=(+/-)1.5
x1=1.5+0.5-----> x1=2
x2=-1.5+0.5---- > x2=-1

for x=2
y=x²----> y=2²----> y=4
the point is (2,4)

for x=-1
y=x²----> y=(-1)²---> y=1
the point is (-1,1)

the answer is
the solution of the system are the points
(2,4) and (-1,1)


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a) v=\frac{d_{tot}}{t_{tot}}=\frac{(3 h)(60 mph)+20 mi}{3 h +t_2}

The average speed is equal to the ratio between the total distance (d_{tot} and the total time taken (t_{tot}):

v=\frac{d_{tot}}{t_{tot}}

the distance travelled by the trucker in the first 3 hour can be written as the time multiplied by the velocity:

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The total time is equal to the first 3 hours + the time taken to cover the following 20 miles in the city:

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b) 0.50 h (half a hour)

Since we know the value of the average speed, v=57.14 mph, we can substitute it into the previous equation to find the value of t_2, the time the trucker drove in the city:

v=\frac{200 mi}{3h +t_2}\\3h+t_2 = \frac{200 mi}{v}\\t_2 = \frac{200 mi}{v}-3h=\frac{200 mi}{57.14 mph}-3 h=0.50 h


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