I need help with this someone please

Mathematics

1 answer:

Alex2 years ago

7 0

3/1 is your slope for that

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H=-16t^2+46t+6 after how many seconds will the ball hit the ground?

ANEK [815]

Ball hit the ground is when the function equals zero
0=-16t^2+46t+6
we can't factor so use quadraticformula

for
ax^2+bx+c=0
x= $\frac{-b+/- \sqrt{b^2-4ac} }{2a}$
x= $\frac{-46+/- \sqrt{46^2-4(-16)(6)} }{2(-16)}$
x= $\frac{-46+/- \sqrt{2116+384} }{-32}$
x= $\frac{-46+/- \sqrt{2500} }{-32}$
x= $\frac{-46+/- 50 }{-32}$
x= $\frac{-46+ 50 }{-32}$ or $\frac{-46- 50 }{-32}$
x= $\frac{4}{-32}$ or x= $\frac{-96}{-32}$
x=-1/6 or x=3

we cannot have negative time because time starts at 0
disregard -1/6 as a possible answer

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sergiy2304 [10]

I think the answer is 21.

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BabaBlast [244]

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Need help I forgot how to do this

Marat540 [252]

Answer:

Step-by-step explanation:

you have to find the discriminant

b²-4ac for each equation

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for your given problems

A. discriminant =(-2)²-4*2*15 will be negative

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oee [108]

The equation y= 2 $(x+1)^2$ has one real root and that is x=-1.

What is real roots of the equation?

We are aware that when we resolve a linear or quadratic equation, we always arrive at the value variable of the equation, or, to put it another way, we always locate the equation's solution. This "solution" is what we refer to as the real roots. For instance, when the equation $X^2$ -7x+12=0 is solved, the actual roots are 3 and 4.