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Butoxors [25]
4 years ago
14

What is the solution to this equation?

Mathematics
1 answer:
zloy xaker [14]4 years ago
6 0
The solution is x=50
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What is the slope of the line that passes
Snowcat [4.5K]

Answer:the answer is 5/2 because if yo subtract y2 from y1 and x2 from y2 thats the answer you get.

But you might not get that answer straight away you need to simplify the initial value and keep going until you cant simplify without a remainder.

Step-by-step explanation:

5 0
4 years ago
Read 2 more answers
PLEASE HELP!!! WILL MARK BRAINLIEST!! THX! A golfer hits a ball with an initial velocity of 32.7 m/s from the ground. Find the f
OLEGan [10]

Answer:

See below

Step-by-step explanation:

<u>First Problem</u>

The ball hits the ground when h(t)=0, therefore:

h(t)=-4.9t^2+v_0t+h_0

0=-4.9t^2+32.7t

0=t(-4.9t+32.7)

t=0 and t=\frac{32.7}{4.9}\approx6.67

Since the ball is in the air before it hits the ground, t=6.67 (seconds) is the more appropriate choice.

<u>Second Problem</u>

The maximum height of the ball is determined when t=-\frac{b}{2a}, therefore:

t=-\frac{b}{2a}

t=-\frac{32.7}{2(-4.9)}

t=-\frac{32.7}{-9.8}

t\approx3.34

This means that the height of the ball is at its maximum after 3.34 seconds:

h(t)=-4.9t^2+32.7t

h(3.34)=-4.9(3.34)^2+32.7(3.34)

h(3.34)\approx54.55

Thus, the answer is 54.55 (meters).

<u>Third Problem</u>

Refer to the second problem

<u>Fourth Problem</u>

<u />h(t)=-4.9t^2+32.7t<u />

<u />h(4.3)=-4.9(4.3)^2+32.7(4.3)<u />

<u />h(4.3)\approx50.01<u />

<u />

Therefore, the height of the ball after 4.3 seconds is 50.01 (meters).

<u>Fifth Problem</u>

The ball will be 24 meters off the ground when h(t)=24, therefore:

h(t)=-4.9t^2+32.7t

24=-4.9t^2+32.7t

0=-4.9t^2+32.7t-24

t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

t=\frac{-32.7\pm\sqrt{(32.7)^2-4(-4.9)(-24)}}{2(-4.9)}

t_1\approx0.84

t_2\approx5.83

Therefore, the ball will be 24 meters off the ground after 0.84 (seconds) and 5.83 (seconds)

5 0
2 years ago
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