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Harlamova29_29 [7]
3 years ago
6

I need help anyoneeeeeee

Mathematics
1 answer:
pshichka [43]3 years ago
8 0

Answer:

answer is 1004.8 cubic feet

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Given f(x) = 6x^4 – 10x^3 + 40x – 50, find f(2)
ASHA 777 [7]
F(x)=6x^4-10x^3+40x-50, plug 2 in for x
f(2)=6(2)^4-10(3)^3+40(2)-50
f(2)=12^4-30^3+80-50
f(2)=20735-27,000+80-50
f(2)=-6,235
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13+(3+7)=(13+3)+7 what property is that​
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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.

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3 years ago
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