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lesantik [10]
2 years ago
8

A golfer hit a golf ball from a tee box that is 6 yards above the ground. The graph shows the height in yards of the golf ball a

bove the ground as a quadratic function of x, the horizontal distance in yards of the golf ball from the tee box. What is the domain of the function for this situation?
Mathematics
1 answer:
Vaselesa [24]2 years ago
4 0

Answer: C. 0 ≤ x ≤ 230

You might be interested in
Write an equation in slope-intercept form for the line that passes through the given point and is perpendicular to the following
Valentin [98]
The perpendicular equation would include a slope that is the opposite reciprocal of the original slope.

Steps:
1. Get x to the other side in the original equation. This making the slope -4 or -4/1.

2. Turn the slope into it’s opposite reciprocal m = 1/4.

3. If you use point-slope form, y - y1 = m( x - x1 ), you can substitute y1 and x1 with the numbers in the point given. But since we previously found the opposite reciprocal, we will replace “m” as well. *By the way, the subtraction of a negative makes a positive. [y + 3 = 1/4( x + 4 )]

4. Solve:
A: Distribute (y + 3 = 1/4x + 1)
B: Subtract 3 from both sides (y = 1/4x -2)

Perpendicular Equation: y = 1/4x - 2
5 0
2 years ago
Read 2 more answers
PLEASE ANSWER THIS ASAP !!!
Inga [223]

Answer:

Step-by-step explanation:

Finding the answer to the second box is easy. Just look at where the line hits the y axis. That point is (0,5). Put a 5 in the second box.

-2

Now pick two points How about (0,5) and (2,1)

Givens

y2 = 5

y1 = 1

x2 = 0

x1 = 2

Formula

m = (y2 - y1) / (x2 - x1)

m = (5 - 1)/(0 - 2)

m = 4 / - 2

m = - 2

Answer

So the first box contains - 2

7 0
2 years ago
Help me to answer this question pl​s
attashe74 [19]

Problem 1

Draw a straight line and plot P anywhere on it. Use the compass to trace out a faint circle of radius 8 cm with center P. This circle crosses the previous line at point Q.

Repeat these steps to set up another circle centered at Q and keep the radius the same. The two circles cross at two locations. Let's mark one of those locations point X. From here, we could connect points X, P, Q to form an equilateral triangle. However, we only want the 60 degree angle from it.

With P as the center, draw another circle with radius 7.5 cm. This circle will cross the ray PX at location R.

Refer to the diagram below.

=====================================================

Problem 2

I'm not sure why your teacher wants you to use a compass and straightedge to construct an 80 degree angle. Such a task is not possible. The proof is lengthy but look up the term "constructible angles" and you'll find that only angles of the form 3n are possible to make with compass/straight edge.

In other words, you can only do multiples of 3. Unfortunately 80 is not a multiple of 3. I used GeoGebra to create the image below, as well as problem 1.

8 0
2 years ago
Can you please help me with#11?I'm not sure how to find the answer.​
almond37 [142]

Answer:

You need to find the x and y intercepts so...

The x intercept is (18,0) since x=18

The y intercept is (0,12) since y=12

Step-by-step explanation:

Cover the y when you are finding the y intercept

Cover the x when you are finding the x intercept

8x/8

144/8

=18

12y/12

144/112

=12

5 0
2 years ago
Helppppppppppppppppppppp!!!!!!!!!1
vitfil [10]

Answer: I need help too

Step-by-step explanation:

I think is -6

6 0
2 years ago
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