In this question, you have to use the Pythagorean theorem. (the formula is in the box, ‘a squared’ and ‘b squared’ is used to find the legs of the triangle, while the ‘c squared’ in the formula is used to find the hypotenuse of the triangle.
From a tee box that is 6 yards above the ground, a golfer hit a ball. The Domain of the function is [0,230].
Given that,
In the picture there is a question with a graph.
From a tee box that is 6 yards above the ground, a golfer hit a ball. The graph displays the height of the golf ball above the ground in yards as a quadratic function of x, the golf ball's horizontal distance from the box in yards.
We have to find the domain of the function in the situation.
The domain is nothing but All of a function's x-values, or inputs, make up the domain, and all of a function's y-values, or outputs, make up the range.
The domain of a graph is every value in the graph, from left to right. The graph's entire range, from lower to higher numbers, represents the range.
Therefore, the Domain of the function is [0,230].
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Answer:
(x +3)^2 + (y -3)^2 = 3^2
Step-by-step explanation:
Your first task is to find the center.
The longest possible line of the circle goes from (-3,0) to (-3,6)
So the center is at the midpoint of these two points.
The midpoint is at (-3 - 3)/2 , (6+ 0)/2 or (-3,3)
That gives you the center of the circle.
So far what you have is
(x +3)^2 + (y -3)^2 = r^2
Now the distance of the center to the bottom point is (-3,3) to (-3,0)
r^2 = (x2 - x1)^2 + (y2 - y1)^2
x2 = - 3
x1 = - 3
y2 = 3
y1 = 0
r^2 = (-3 - -3)^2 + (3 - 0)^2
r^2 = 3^2
The entire answer is
(x +3)^2 + (y -3)^2 = 3^2
Answer:
The answer is 76.9 rounded or rounds.
Step-by-step explanation:
Solution
Given that:
Mean = μ = 72.5
Standard deviation = σ = 6.5
Now
By applying the standard normal table we have the following:
P(Z < z) = 75%
= P(Z < z)
= 0.75
Thus
z =0.67 by using the standard normal z table
Applying the z-score formula we have:
x = z * σ + μ
x = 0.67*6.5+72.5
x= 76.855 ≈ 76.9
Therefore the cutoff time is 76.9 rounds.
