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3 years ago

Help with algebra pls help

Mathematics

1 answer:

EastWind [94]3 years ago

8 0

9514 1404 393

Answer:

a. 1.48 seconds

Step-by-step explanation:

You want to find the larger value of t such that h(t) = 10.

-16t^2 +25t +8 = 10

16t^2 -25t +2 = 0 . . . . subtract the left side to get standard form

Using the quadratic formula, we find the values of t to be ...

t = (-(-25) ± √((-25)^2 -4(16)(2)))/(2(16)) = (25±√497)/32

t ≈ 0.08 or 1.48

The ball goes in the hoop about 1.48 seconds after it is thrown.

<em>Additional comment</em>

The quadratic formula tells us the solution to ...

ax² +bx +c = 0

is given by ...

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here, we have a=16, b=-25, c=2. Of course, our variable is t, not x, but the relation is the same.

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Step-by-step explanation:

In the given question the numeric value is missing which is defined in the attached file please fine it.

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$\to f(x) = 0.19\ for \ x=14\\\\\to f(x) = 0.29 \ for\ x=7\\\\\to f(x) = 0.38\ for \ x=1\\\\\to f(x)=0.14 \ for \ x=0\\$

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