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3 years ago

5

How to solve and the steps on how to do it because it is hard for me

1 answer:

spayn [35]3 years ago

6 0

For number one:
u first subtract 6 from 24
24-6= 18

then u devide 18 by 3
18/3= 6 miles

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at a party, everyone shook hands with everybody else. there were 66 handshakes. How many people shook hands?

Whitepunk [10]

Answer:

12

Step-by-step explanation:

0 + 1 + 2 + 3+ 4 + 5 + 6 + 7 +8 + 9 + 10 + 11 seems to equal 66. That means there are 11 + 1 = 12 people (don’t forget to count the person who shakes 0 hands)

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3 years ago

A binder normally costs $8.99. use this information to answer the questions below

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You are asking a question that cannot be solved by anyone outside of your school.

Step-by-step explanation:

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3 years ago

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3 years ago

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