Type the correct answer in each box. Use numerals instead of words. If necessary, round your answer to the nearest hundredth.
Ana and Taylor are on the lacrosse team and practicing their passes. The vertical height, a(x), of Ana’s pass x feet from where it was thrown is modeled in this table.
0 10 20 30 40 50 60
0 20 32 36 32 20 0
The vertical height, t(x), of Taylor’s pass x feet from where it was thrown is modeled by this equation.
t(x) = -0.05 (x^2 - 50x)
Complete the statements comparing their passes.
The difference of the maximum heights is ___ feet.
The difference of the total distances traveled is ___ feet.
To find the maximum height of Ana's pass, we look for the highest value in the height column. From the table, we can see that the maximum height of Ana's pass is 36 feet.
To find the maximum height of Taylor's pass, we need to find the vertex of the parabola defined by the equation t(x) = -0.05(x^2 - 50x). The x-coordinate of the vertex can be found using the formula x = -b/2a, where a = -0.05 and b = -50. Plugging in these values, we get x = -(-50)/(2*-0.05) = 500/0.1 = 5000.
Plugging x = 5000 into the equation t(x) = -0.05(x^2 - 50x), we can find the maximum height of Taylor's pass:
t(5000) = -0.05(5000^2 - 50*5000) = -0.05(25,000,000 - 250,000) = -0.05(24,750,000) = -1,237,500.
Therefore, the maximum height of Taylor's pass is -1,237,500 feet.
The difference of the maximum heights is 36 - (-1,237,500) = 1,237,536 feet.
To find the total distance traveled by Ana's pass, we need to find the difference between the first and last x values in the table. In this case, the first x value is 0 and the last x value is 60. Therefore, the total distance traveled by Ana's pass is 60 - 0 = 60 feet.
To find the total distance traveled by Taylor's pass, we need to find the x-intercepts of the parabola defined by the equation t(x) = -0.05(x^2 - 50x). The x-intercepts can be found by setting t(x) = 0 and solving for x:
-0.05(x^2 - 50x) = 0
x^2 - 50x = 0
x(x - 50) = 0
x = 0 or x = 50.
Therefore, the total distance traveled by Taylor's pass is 0 - 50 = -50 feet.
The difference of the total distances traveled is 60 - (-50) = 110 feet.
Very quickly
t(x) = -0.05 (x^2 - 50x)
where is Taylor height max?
d/dx (x^2-50) = 0
2 x = 50
x = 25 for max
and that max is
-0.05 ( 625 - 50) = 0.05 (575 ) = 28.75
which is NOT -1,237,500 feet
repair needed desperately !
Apologies for the mistake in the previous response. Let's correct it.
To find the maximum height of Taylor's pass, we need to find the vertex of the parabola defined by the equation t(x) = -0.05(x^2 - 50x). The x-coordinate of the vertex can be found using the formula x = -b/2a, where a = -0.05 and b = -50. Plugging in these values, we get x = -(-50)/(2*(-0.05)) = 50/0.1 = 500.
Plugging x = 500 into the equation t(x) = -0.05(x^2 - 50x), we can find the maximum height of Taylor's pass:
t(500) = -0.05(500^2 - 50*500) = -0.05(250,000 - 25,000) = -0.05(225,000) = -11,250.
Therefore, the maximum height of Taylor's pass is -11,250 feet.
The difference of the maximum heights is 36 - (-11,250) = 11,286 feet.
To find the total distance traveled by Ana's pass, we need to find the difference between the first and last x values in the table. In this case, the first x value is 0 and the last x value is 60. Therefore, the total distance traveled by Ana's pass is 60 - 0 = 60 feet.
To find the total distance traveled by Taylor's pass, we can use the same x values. Plugging in x = 0 into the equation t(x) = -0.05(x^2 - 50x), we get t(0) = -0.05(0^2 - 50*0) = 0. Therefore, the starting point of Taylor's pass is at x = 0.
Plugging in x = 60 into the equation t(x) = -0.05(x^2 - 50x), we get t(60) = -0.05(60^2 - 50*60) = -0.05(3,600 - 3,000) = -0.05(600) = -30.
Therefore, the total distance traveled by Taylor's pass is 0 - (-30) = 30 feet.
The difference of the total distances traveled is 60 - 30 = 30 feet.
To find the total distance traveled by Taylor's pass, we can use the same x values. Plugging in x = 0 into the equation t(x) = -0.05(x^2 - 50x), we get t(0) = -0.05(0^2 - 50*0) = 0. Therefore, the starting point of Taylor's pass is at x = 0.
yes,Taylor pass starts at x = 09
where does it end?
It ends when x^2 - 50 x = 0 again
x (x-50) = 0
x = 50 <==================
So it went 50 feet
How high did it go
if x = 25, the vertex
t = -0.05(x^2 - 50x) = -.05 (625 - 1250) = -.05 ( -625) = + 31.25
Apologies for the confusion caused by the previous response. You are correct.
Taylor's pass starts at x = 0 and ends at x = 50 feet. Therefore, the total distance traveled by Taylor's pass is 50 feet.
At the x-coordinate of the vertex (x = 25), the height of Taylor's pass is given by t(x) = -0.05(x^2 - 50x). Plugging in x = 25, we get:
t(25) = -0.05(25^2 - 50*25) = -0.05(625 - 1250) = -0.05(-625) = +31.25.
Therefore, the maximum height of Taylor's pass is 31.25 feet.