A long jumper leaves the ground at an angle of 20 o above the horizontal at a speed of11 m/sec. The height of the jumper can be modeled by h x = − 0.046 x 2 + 0.364 x , where h is the jumper’s height in meters and x is the horizontal distance from the point of launch. a. At what horizontal distance from the point of launch does the maximum height occur? Round to 2 decimal places. b. What is the maximum height of the long jumper? Round to 2 decimal places. c. What is the length of the jump? Round to 1 decimal place.
A long jumper leaves the ground at an angle of 20 o above the horizontal at a speed of11 m/sec. The height of the jumper can be modeled by h x = − 0.046 x 2 + 0.364 x , where h is the jumper’s height in meters and x is the horizontal distance from the point of launch. a. At what horizontal distance from the point of launch does the maximum height occur? Round to 2 decimal places. b. What is the maximum height of the long jumper? Round to 2 decimal places. c. What is the length of the jump? Round to 1 decimal place.
Solution Summary: The author calculates the horizontal distance from the point of launch at which the maximum height of the jumper occurs using f(x)=-0.046x2+0.364x.
A long jumper leaves the ground at an angle of
20
o
above the horizontal at a speed of11 m/sec. The height of the jumper can be modeled by
h
x
=
−
0.046
x
2
+
0.364
x
,
where h is the jumper’s height in meters and x is the horizontal distance from the point of launch.
a. At what horizontal distance from the point of launch does the maximum height occur? Round to 2 decimal places.
b. What is the maximum height of the long jumper? Round to 2 decimal places.
c. What is the length of the jump? Round to 1 decimal place.
Write the given third order linear equation as an equivalent system of first order equations with initial values.
Use
Y1 = Y, Y2 = y', and y3 = y".
-
-
√ (3t¹ + 3 − t³)y" — y" + (3t² + 3)y' + (3t — 3t¹) y = 1 − 3t²
\y(3) = 1, y′(3) = −2, y″(3) = −3
(8) - (888) -
with initial values
Y
=
If you don't get this in 3 tries, you can get a hint.
Question 2
1 pts
Let A be the value of the triple integral
SSS.
(x³ y² z) dV where D is the region
D
bounded by the planes 3z + 5y = 15, 4z — 5y = 20, x = 0, x = 1, and z = 0.
Then the value of sin(3A) is
-0.003
0.496
-0.408
-0.420
0.384
-0.162
0.367
0.364
Question 1
Let A be the value of the triple integral SSS₂ (x + 22)
=
1 pts
dV where D is the
region in
0, y = 2, y = 2x, z = 0, and
the first octant bounded by the planes x
z = 1 + 2x + y. Then the value of cos(A/4) is
-0.411
0.709
0.067
-0.841
0.578
-0.913
-0.908
-0.120
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