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34kurt
3 years ago
5

The object will hit the ground after 5 seconds. You can rewrite the quadratic function as a quadratic equation set equal to zero

to find the zeros of the function 0 = –16t2 + 80t + 0. You can factor or use the quadratic formula to get t = 0 and t = 5. Therefore, it is on the ground at t = 0 (time of launch) and then hits the ground at t = 5 seconds
Mathematics
1 answer:
mario62 [17]3 years ago
8 0

Answer:

289

Step-by-step explanation:

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Solve -3(-8b+7)=3(2b-1)
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Answer: b=1

Step-by-step explanation:

Given the equation -3(-8b+7)=3(2b-1) you need to solve for the variable "b".

You need to apply Distributive property on both sides of the equation:

-3(-8b+7)=3(2b-1)\\\\24b-21=6b-3

Now you need to add 21 to both sides of the equation:

24b-21+21=6b-3+21\\\\24b=6b+18

Subtract 6b from both sides:

24b-6b=6b+18-6b\\\\18b=18

And finally, you can divide both sides of the equation by 18:

\frac{18b}{18}=\frac{18}{18}\\\\b=1

4 0
2 years ago
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5 0
3 years ago
5x +6y = 20 (1)
slega [8]
(2) + (2)

=> 8x - 6y = -46 ———(3)

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=> 6y = 20 + 10

=> 6y = 30

=> y = 5

x = -2, y = 5 => C





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