Which trigonometric function requires a domain restriction of [0,pi] to make it invertable?
a. f(x)= sin x
b. f(x)= cos x*******
c. f(x)= tan x
d. f(x)= csc x
e. f(x)= sec x
f. f(x)= cot x
Which of the following points could be the initial point of vector v if it has a magnitude of 10 an the terminal point (-1,5)?
a. (-0.1,0.5)
b. (2,4)
c. (-11,-5)*******
d. (-7,-3)
A baseball player hits a ball at an angle of 52 degree and at a height of 4.8 ft. If the ball's initial velocity after being hit is 150 ft./s and if no one catches the ball, when will it hit the ground? Remember that the acceleration due to gravity is 32 ft./s^2
a. 5.77 seconds
b. 5.82 seconds
c. 7.43 seconds*******
d. 7.39 seconds
so uhhhhhhh what are the correct answers?
a. f(x)= sin x
Why did the trigonometric function go to therapy? It had issues with its identity!
c. (-11,-5)
Why don't mathematicians like the number 7? Because it's odd!
c. 7.43 seconds
Why did the baseball player bring a ladder to the game? He wanted to hit a home run!
1. The trigonometric function that requires a domain restriction of [0,pi] to make it invertible is b. f(x) = cos x.
2. The initial point of vector v, with a magnitude of 10 and terminal point (-1,5), can be calculated using the formula:
(x, y) = (terminal point - magnitude * cos θ, magnitude * sin θ)
where θ is the angle between the positive x-axis and the vector v.
Calculating for option c. (-11, -5):
x = -1 - 10 * cos θ = -1 - 10 * cos(arctan(5/-11)) = -1 - 10 * cos(-0.4237) ≈ -11.118
y = 10 * sin θ = 10 * sin(arctan(5/-11)) = 10 * sin(-0.4237) ≈ -4.995
So, option c. (-11, -5) can be the initial point of vector v.
3. To find when the ball will hit the ground, we need to determine the time it takes for the ball to reach a height of 0 ft. The equation for the height of the ball as a function of time can be written as:
h(t) = h0 + v0 * sin(θ) * t - (1/2) * g * t^2
Setting h(t) = 0 and solving for t:
0 = 4.8 + 150 * sin(52°) * t - (1/2) * 32 * t^2
0 = 4.8 + 114.79 * t - 16 * t^2
Solving this quadratic equation, we find that t ≈ 7.43 seconds.
Therefore, the answer is c. 7.43 seconds.
1. To find the trigonometric function that requires a domain restriction of [0, pi] to make it invertible, we need to look at the definitions of the trigonometric functions and their domains.
a. f(x) = sin x: The domain of the sine function is all real numbers, so it does not require a domain restriction of [0, pi] to make it invertible.
b. f(x) = cos x: The domain of the cosine function is all real numbers, so it does not require a domain restriction of [0, pi] to make it invertible.
c. f(x) = tan x: The tangent function has vertical asymptotes at x = (2n + 1)(pi/2), where n is an integer. This means that it is undefined at these points, so it does require a domain restriction to be invertible. However, the domain restriction should be (-pi/2, pi/2) for tan x to be invertible, not [0, pi].
d. f(x) = csc x: The cosecant function has vertical asymptotes at x = n(pi), where n is an integer. This means that it is undefined at these points, so it does require a domain restriction to be invertible. However, the domain restriction should be (-infinity, -pi) U (0, infinity) for csc x to be invertible, not [0, pi].
e. f(x) = sec x: The secant function has vertical asymptotes at x = (2n + 1)(pi/2), where n is an integer. This means that it is undefined at these points, so it does require a domain restriction to be invertible. However, the domain restriction should be (-pi/2, 0) U (pi/2, infinity), not [0, pi] for sec x to be invertible.
f. f(x) = cot x: The cotangent function has vertical asymptotes at x = n(pi), where n is an integer. This means that it is undefined at these points, so it does require a domain restriction to be invertible. In this case, the domain restriction of [0, pi] is necessary for cot x to be invertible.
Therefore, the correct answer is b. f(x) = cos x.
2. To find the initial point of a vector with a magnitude of 10 and terminal point (-1, 5), we can use the vector magnitude formula.
The vector magnitude formula is:
Magnitude = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Let's calculate the magnitude using the given values:
10 = sqrt(((-1) - x1)^2 + (5 - y1)^2)
Squaring both sides:
100 = ((-1) - x1)^2 + (5 - y1)^2
Plugging in the options:
a. (-0.1, 0.5): 100 = ((-1) - (-0.1))^2 + (5 - 0.5)^2
b. (2, 4): 100 = ((-1) - 2)^2 + (5 - 4)^2
c. (-11, -5): 100 = ((-1) - (-11))^2 + (5 - (-5))^2
d. (-7, -3): 100 = ((-1) - (-7))^2 + (5 - (-3))^2
After calculating the values, we find that option c. (-11, -5) satisfies the equation.
Therefore, the correct answer is c. (-11, -5).
3. To find when the baseball hits the ground after being hit at a certain angle and height, we can use the kinematic equation for vertical motion:
y = y0 + v0y * t - (1/2) * g * t^2
Where:
- y is the vertical position of the ball at time t,
- y0 is the initial vertical position of the ball,
- v0y is the vertical component of the initial velocity of the ball,
- g is the acceleration due to gravity, and
- t is the time.
In this case, we are given:
- y0 = 4.8 ft (the height at which the ball was hit),
- v0y = v0 * sin(theta) (the vertical component of the initial velocity, where v0 is the initial velocity and theta is the angle at which the ball was hit),
- g = 32 ft/s^2 (the acceleration due to gravity).
Substituting the given values into the equation, we have:
0 = 4.8 + (150 * sin(52)) * t - (1/2) * 32 * t^2
Simplifying and rearranging the equation:
0 = 4.8 + 78t - 16t^2
Rearranging to a quadratic equation form:
16t^2 - 78t - 4.8 = 0
Using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
Where:
- a = 16,
- b = -78, and
- c = -4.8.
After substituting the values into the formula and solving, we find:
t ≈ 7.43 seconds
Therefore, the correct answer is c. 7.43 seconds.