Let the original length of the land be L and the original width be W.
Given that the original area is 200 x 300, we have:
L * W = 200 * 300
L * W = 60000
After adding x meters to the length and width, the new length will be L + x and the new width will be W + x.
The new area will be:
(L + x) * (W + x)
Given that the new area is double the original area, we have:
2 * L * W = (L + x) * (W + x)
2 * 60000 = (200 + x) * (300 + x)
120000 = (200 + x) * (300 + x)
Expanding the right side:
120000 = 60000 + 200x + 300x + x^2
120000 = 60000 + 500x + x^2
0 = x^2 + 500x - 60000
Now, we need to solve this quadratic equation. By factoring or using the quadratic formula, we find that x = 60 meters or x = -560 meters.
Since length and width cannot have negative values, x = 60 meters.
A plot of land doubles in size by adding x meters to the length and x meters to the width of the land.
If the original plot had an area of 200 by 300 meters, what is the value of x?
3 answers
The CHS Baseball team was on the field and the batter popped the ball up. The equation b(t) = 80t - 16t^2 + 3.5 represents the height of the ball above the ground in feet as a function of time in seconds. How long will the catcher have to get in position to catch the ball before it hits the ground? Round to the nearest second. blank seconds
To find the time it takes for the ball to hit the ground, we need to find when the height of the ball is equal to 0.
So, set b(t) = 0:
0 = 80t - 16t^2 + 3.5
Rearrange the equation:
16t^2 - 80t + 3.5 = 0
Now, we need to solve this quadratic equation. We can use the quadratic formula:
t = (-(-80) ± sqrt((-80)^2 - 4*16*3.5)) / (2*16)
t = (80 ± sqrt(6400 - 224))/32
t = (80 ± sqrt(6176))/32
t ≈ (80 ± 78.5)/32
Therefore, t ≈ (80 + 78.5)/32 or t ≈ (80 - 78.5)/32
t ≈ 158.5/32 or t ≈ 1.5/32
t ≈ 4.953125 seconds or t ≈ 0.046875 seconds
Since time cannot be negative, the catcher will have approximately 5 seconds to get in position to catch the ball before it hits the ground.
So, set b(t) = 0:
0 = 80t - 16t^2 + 3.5
Rearrange the equation:
16t^2 - 80t + 3.5 = 0
Now, we need to solve this quadratic equation. We can use the quadratic formula:
t = (-(-80) ± sqrt((-80)^2 - 4*16*3.5)) / (2*16)
t = (80 ± sqrt(6400 - 224))/32
t = (80 ± sqrt(6176))/32
t ≈ (80 ± 78.5)/32
Therefore, t ≈ (80 + 78.5)/32 or t ≈ (80 - 78.5)/32
t ≈ 158.5/32 or t ≈ 1.5/32
t ≈ 4.953125 seconds or t ≈ 0.046875 seconds
Since time cannot be negative, the catcher will have approximately 5 seconds to get in position to catch the ball before it hits the ground.