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4 years ago

Can you help me with this problem I can't find it out

Mathematics

2 answers:

AnnZ [28]4 years ago

8 0

He estimates that he will earn $44,800.

12% of 40,000 is 4,800
40,000+4,800=44,800

zhenek [66]4 years ago

5 0

Money earned last year = $40,000
Money estimated to be earned this year = $40,000 + 12% of $40,000
= $40,000 + (12/100 × $40,000)
= $40,000 + $4,800
= $44,800
Plz mark it brainliest

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Answer:

See below

Step-by-step explanation:

First Problem

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

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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:

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

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$h(3.34)\approx54.55$

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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Answer:

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The general solution becomes:

$y(t) = A {e}^{ - 2t} \cos(t) + B {e}^{ - 2t} \sin(t)$

We differentiate to get:

$y'(t) = - A {e}^{ - 2t} \sin(t) - 2 A {e}^{ - 2t} \cos(t) + B {e}^{ - 2t} \cos(t) - 2B {e}^{ - 2t} \sin(t)$

We apply the initial conditions to get;

$y( \frac{\pi}{2} ) = A {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) + B {e}^{ - 2\pi} \sin( \frac{\pi}{2} )$

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Also;

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