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a_sh-v [17]
3 years ago
7

Please help me AAAAAA

Mathematics
1 answer:
bezimeni [28]3 years ago
8 0

Answer:

45

Step-by-step explanation:

AEF is a similar triangle to ABC. that means it has the same angles, and the sides (and all other lines in the triangle) are scaled from the ABC length to the AEF length by the same factor f.

now, what is f ?

we know this from the relation of AC to FA.

FA = 12 mm

AC = 12 + 28 = 40 mm

so, going from AC to FA we multiply AC by f so that

AC × f = FA

40 × f = 12

f = 12/40 = 3/10

all other sides, heights, ... if ABC translate to their smaller counterparts in AEF by that multiplication with f (= 3/10).

the area of a triangle is

baseline × height / 2

aABC = 500

and because of the similarity we don't need to calculate the side and height in absolute numbers. we can use the relative sizes by referring to the original dimensions and the scaling factor f.

baseline small = baseline large × f

height small = height large × f

we know that

baseline large × height large / 2 = 500

baseline large × height large = 1000

aAEF = baseline small × height small / 2 =

= baseline large × f × height large × f / 2 =

= baseline large × height large × f² / 2 =

= 1000 × f² / 2 = 500 × f² = 500 ×(3/10)² =

= 500 × 9/100 = 5 × 9 = 45 mm²

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Total Time = 4.51 s

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Solution:

- It firstly asks you to prove that that statement is true. To prove it, we will need a little bit of kinematics:

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- We know that ball rebounds back to 7/8 of its previous height h. So we will calculate times for each bounce:

1st : 0.25*\sqrt{15}\\\\2nd: 0.25*\sqrt{15} + 0.25*\sqrt{15*\frac{7}{8} } + 0.25*\sqrt{15*\frac{7}{8} } = 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} }\\\\3rd: 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 2*0.25*\sqrt{15*(\frac{7}{8} })^2\\\\= 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2\\\\4th: 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2 + 2*0.25*\sqrt{15*(\frac{7}{8} })^3 \\\\

= 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2 + 0.5*\sqrt{15*(\frac{7}{8} })^3

- How long it has been bouncing at nth bounce, we will look at the pattern between 1st, 2nd and 3rd and 4th bounce times calculated above. We see it follows a geometric series with formula:

  Total Time ( nth bounce ) = Sum to nth ( \frac{1}{2}*\sqrt{15*(\frac{7}{8})^(^i^-^1^) }  - \frac{1}{4}*\sqrt{15})

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