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77julia77 [94]
3 years ago
8

Solve each of these equations. Explain or show your reasoning.

Mathematics
1 answer:
Svetlanka [38]3 years ago
4 0

1.

2(x+5)=3x+1 \\2x+10=3x+1 \\-x=-9 \\x=9

2.

3y-4=6-2y \\5y=10 \\y=2

3.

3(n+2)=9(6-n) \\3n+6=54-9n \\12n=48 \\n=\frac{48}{12}=4

Hope this helps.

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In order to evaluate this integral, we need to write this rational function in a simpler way:
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</span>= - \int { \frac{1}{P} \, dP +&#10;6\int { \frac{0.2}{0.2P-400} } \, dP<span>
</span>= - ln |P| + 6 ln |0.2P - 400| + C<span>

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</span>0 = - ln |10000| + 6 ln |0.2(10000) - <span>400| + C
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The height H of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 9
mariarad [96]

Answer:

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

Step-by-step explanation:

Statement is incorrect. Correct form is presented below:

<em>The height </em>h(t)<em> of an ball that is thrown straight upward from an initial position 3 feet off the ground with initial velocity of 90 feet per second is given by equation </em>h(t) = 3 +90\cdot t -16\cdot t^{2}<em>, where </em>t<em> is time in seconds. After how many seconds will the ball be 84 feet above the ground. </em>

We equalize the kinematic formula to 84 feet and solve the resulting second-order polynomial by Quadratic Formula to determine the instants associated with such height:

3+90\cdot t -16\cdot t^{2} = 84

16\cdot t^{2}-90\cdot t +81 = 0 (1)

By Quadratic Formula:

t_{1,2} = \frac{90\pm \sqrt{(-90)^{2}-4\cdot (16)\cdot (81)}}{2\cdot (16)}

t_{1} = 4.5\,s, t_{2} = 1.125\,s

The ball will be 84 feet above the ground 1.125 seconds and 4.5 seconds after launch.

3 0
3 years ago
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