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Ghella [55]
3 years ago
9

An arrow is shot vertically upward from a platform 41ft41⁢ft high at a rate of 155ft/sec155⁢ft/sec. when will the arrow hit the

ground? use the formula: h=−16t2+
Mathematics
1 answer:
lys-0071 [83]3 years ago
7 0
H(t)=-16t²+155t+41
0=-16t²+155t+41
16t²-155t-41=0
Using the quadratic formula, we get a positive value for t of  <span>9.94516294765 seconds as the time the arrow is in the air
☺☺☺☺
</span>

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Answer:

Step-by-step explanation:

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f(-9+h)-f(-9)=h²-18h+81 -81 because : f(-9) = (-9)² = 81

f(-9+h)-f(-9)=h²-18h

(f(-9+h)-f(-9))/h=(h²-18h)/h = h(h-18)/h =h-18

lim  (f(-9+h)-f(-9))/h = lim(h-18= - 18

h→0                             h→0

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Double Points! (i have no clue how to do this)<br> Find the value of the variables in the figure.
Aleks [24]

Answer:

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Step-by-step explanation:

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Subtract 3x from both sides:

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Divide each side by 2:

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Now we can see that ( 5x - 7 ) and ( 4y + 3 ) are supplementary angles, which means they will add up to 180 degrees:

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Substitute 12 in for x to solve for y:

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6 0
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Answer:

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