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Mathematics

1 answer:

mariarad [96]3 years ago

8 0

I’m a bit confused by the choices you have been given. Are there any other options? If not, I would choose (5.3,9.8) as 9.8 is correct

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According to this diagram, what is tan 62 degrees ?

sammy [17]

Check the picture below

in case you want to know how to get who's the adjacent or opposite, notice in the picture, if you put your eye on the angle itself, what you'd be facing is the opposite side, the adjacent is the side touching the angle.

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dexar [7]

Here's our equation.

$h=-16t+64t+3$

We want to find out when it returns to ground level (h = 0)

To find this out, we can plug in 0 and solve for t.

$0 = -16t+64t+3 \\ 16t-64t-3=0 \\ use\ the\ quadratic\ formula\ \frac{-b\±\sqrt{b^2-4ac}}{2a} \\ \frac{-(-64)\±\sqrt{(-64)^2-4(16)(-3)}}{2*16} = \frac{64\±\sqrt{4096+192}}{32}$

$= \frac{64\±\sqrt{4288}}{32} = \frac{64\±8\sqrt{67}}{32} = \frac{8\±\sqrt{67}}{4} = \boxed{\frac{8+\sqrt{67}}{4}\ or\ 2-\frac{\sqrt{67}}{4}}$

So the ball will return to the ground at the positive value of $\boxed{\frac{8+\sqrt{67}}{4}}$ seconds.

What about the vertex? Simple! Since all parabolas are symmetrical, we can just take the average between our two answers from above to find t at the vertex and then plug it in to find h!

$\frac{1}2(\frac{8+\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(2+\frac{\sqrt{67}}{4}+2-\frac{\sqrt{67}}{4}) = \frac{1}2(4) = 2$

$h=-16t^2+64t+3 \\ h=-16(2)^2+64(2)+3 \\ \boxed{h=67}$

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Law Incorporation [45]

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