Why should you check the solution in both equations of a system?
1 answer:
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Answer: Anything between 0 and 10, excluding both endpoints.
In terms of symbols we can say 0 < w < 10 where w is the width.
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Explanation:
You could do this with two variables, but I think it's easier to instead use one variable only. This is because the length is dependent on what you pick for the width.
w = width
2w = twice the width
2w-5 = five less than twice the width = length
So,
- width = w
- length = 2w-5
which lead to
area = length*width
area = (2w-5)*w
area = 2w^2-5w
area < 150
2w^2 - 5w < 150
2w^2 - 5w - 150 < 0
To solve this inequality, we will solve the equation 2w^2-5w-150 = 0
Use the quadratic formula. Plug in a = 2, b = -5, c = -150
Ignore the negative solution as it makes no sense to have a negative width.
The only practical root is w = 10.
If w = 10 feet, then the area = 2w^2-5w results in 150 square feet.
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Based on that root, we need to try a sample value that is to the left of it.
Let's say we try w = 5.
2w^2 - 5w < 150
2*5^2 - 5*5 < 150
25 < 150 ... which is true
This shows that if 0 < w < 10, then 2w^2-5w < 150 is true.
Now try something to the right of 10. I'll pick w = 15
2w^2 - 5w < 150
2*15^2 - 5*15 < 150
375 < 150 ... which is false
It means w > 10 leads to 2w^2-5w < 150 being false.
Therefore w > 10 isn't allowed if we want 2w^2-5w < 150 to be true.
Answer:
The time taken for the flare to hit the ground is approximately 10.7 seconds.
Step-by-step explanation:
Given : Suppose a flare is shot from the top of a 120 foot building at a speed of 160 feet per second. The equation models the h height at t seconds of the flare.
To find : How long will it take for the flare to hit the ground?
Solution :
The equation models the h height at t seconds of the flare.
The flare to hit the ground when h=0.
Substitute in the equation,
Applying quadratic formula,
Where, a=-16, b=160 and c=120
Reject the negative value.
Therefore, the time taken for the flare to hit the ground is approximately 10.7 seconds.