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3 years ago

17+4h+2=1−5h solve for h

Mathematics

2 answers:

MAVERICK [17]3 years ago

8 0

Answer:(my bad i did it a different way and i am doing algebra so i though of it as something else my bad)

1-19/4-5

Step-by-step explanation:

17+4h+2=1-5h

17+4h-5h+2=1

19+4h-5h=1

4h-5h=1-19

factor the 4h and 5h

h(4-5)=1-19

1-19/4-5

LUCKY_DIMON [66]3 years ago

4 0

Solution:
17+4h+2 = 1-5h

4h+5h = 1-2-17
9h = -18
h = -2

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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:

First Problem

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.

Second Problem

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).

Third Problem

Refer to the second problem

Fourth Problem

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

$h(4.3)=-4.9(4.3)^2+32.7(4.3)$

$h(4.3)\approx50.01$

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

Fifth Problem

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)

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2 years ago

kkurt [141]

Answer:

$88.

Step-by-step explanation:

Do 22 * 4, your answer should be 88.

7 0

3 years ago