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Gnoma [55]
3 years ago
9

John is standing on a set of bleachers and throws a baseball into the air. the baseball lands on the field. the height of the ba

ll, in meters, is modeled by the function shown in the graph. what's the average rate of change of the height of the ball from the point when John throws it to its maximum height? ​

Mathematics
1 answer:
Aleonysh [2.5K]3 years ago
7 0

Answer:

2 m/s

Step-by-step explanation:

Given the graph of the baseball thrown into air.

John throws the baseball into the air from a given height and then the baseball reaches a maximum height and then comes back to ground.

The height of the baseball (in metres) is modeled by the function graph with respect to time.

To find:

The average rate of change of height of the ball from the point the ball is thrown to the maximum height.

Solution:

Here, the average rate of change can be found by dividing the distance between the two points with the change in time.

The starting time is 0 seconds.

The starting height is 3 metres.

The ending time is 2 seconds i.e. at which the ball reaches the maximum height.

The height is 7 metres.

Therefore, the average rate of change is:

\dfrac{\text{Change in height}}{\text{Change in time}}\\\Rightarrow \dfrac{7-3}{2-0}\\\Rightarrow \dfrac{4}{2}\\\Rightarrow 2\ m/s

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PLEASE HELP ME WITH THIS
Vedmedyk [2.9K]
Let's start with the drawing. (See the drawing at the bottom of this answer.)

Draw a vertical segment.
At the bottom endpoint, draw a horizontal segment to the right.
The angle at the bottom left is a right angle.
So far this should look like an "L" shape.
The vertical segment is the wall, and the horizontal segment is the ground.

Now draw a segment that connects the right endpoint of the horizontal segment and the top endpoint of the vertical segment. Now you have a right triangle. The diagonal segment represents the ladder. The diagonal segment is the hypotenuse. Label the diagonal segment, the hypotenuse, 12 ft.
Label the vertical segment, the wall, 11.8 ft.

The angle at the bottom right is A. It is the angle the ladder makes with the ground. This angle cannot be greater than 75 degrees.

Now we use trigonometry to find the measure of angle A.

For this right triangle, and its angle A, you have a hypotenuse that measures 12 ft, and an opposite leg that measures 11.8 ft.
We need to find angle A.
The trig ratio that relates the opposite leg and the hypotenuse is the sine.

\sin A = \dfrac{opp}{hyp}

\sin A = \dfrac{11.8}{12}

\sin A = 0.98333

Since the sine of angle A equals 0.98333, we use the inverse sine function to find the measure of angle A.

A = \sin^{-1} 0.98333

A = 79.5^\circ

Answer:
The angle the ladder makes with the ground is 79.5 degrees which is greater than 75 degrees, so the ladder it will be unsafe in this position.


           |\
           |  \
           |    \
           |      \
opp = |         \ hyp = 12
= 11.8  |           \
           |             \
           |_______\  A
3 0
3 years ago
Jose unlocks his cellphone by placing his right thumb on a square of 1 centimeter by 1 centimeter at the center of the screen. U
babunello [35]

Answer:

Step-by-step explanation:

We can solve this question using proportion

Perimeter of the square/Time is takes to unlock

We are told in the question that:

The cellphone unlocks when the perimeter reaches 32 centimeters, taking a total of 2.5 seconds.

Hence, the perimeter of the square after 1.5 seconds is:.x = unknown

32/2.5 = x/1.5

Cross Multiply

2.5x = 32 × 1.5

x = 32 × 1.5/2.5

x = 19.2 centimeters

4 0
3 years ago
Write the equation of the line that passes through the given points. (0,4) (-2,-5)
valentina_108 [34]

Answer:

y=-9/2x plus 4

Step-by-step explanation:

y=ax plus b

Use points from the task.

4=0x plus b

so b=4

-5=-2a plus 4

-2a = -9

a = -9/2

y=-9/2x plus 4

8 0
3 years ago
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malfutka [58]

Answer:

y = -34

Step-by-step explanation:

Step 1

1/4 y + 9 = 1/2

1/4 + 9 - 9 = 1/2 - 9

1/4 y = -17/2

Step 2

4 * (1/4 y) = 4* (-17/2)

y = -34

Hope this helps!!

7 0
3 years ago
Help me please help help help​
jolli1 [7]
What is your question?
5 0
3 years ago
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