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slega [8]
1 year ago
9

Y = 7x (6/7,-1/4) ….

Mathematics
1 answer:
oksano4ka [1.4K]1 year ago
7 0

Answer:

This point is not on the line, NOT TRUE

Step-by-step explanation:

The equation of the line is written in the slope-intercept form,

y = mx + b

where m represents the slope

           b represents the y-intercept

where the format of a point is (x, y)

<u>Given:</u> y = 7x (6/7, -1/4)

we can plug in the point (6/7, -1/4)

-1/4 = 7(6/7)

-1/4 = 6

<em>-1/4 does not equal 6 so this point is not on the line</em>

Learn more about slope and y-intercept here: brainly.com/question/12763756

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sweet [91]
Converting to polar coordinates gives

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3 years ago
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dexar [7]
Here's our equation.

h=-16t+64t+3

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

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= \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}


8 0
3 years ago
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denpristay [2]

Answer:

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3 years ago
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Answer:

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have a good day friend :)

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