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3 years ago

The figure below shows concentric circles, both centered at 0.0

Mathematics

1 answer:

Rama09 [41]3 years ago

7 0

Answer:

C. 18 cm

Step-by-step explanation:

The ratio of the sides of the triangle shown is 12 : 15 = 4 : 5. We know it is a right triangle, so we know the missing side length completes the ratio

3 : 4 : 5 = 9 : 12 : 15

Half of XY is 9 cm, so the length of the entire chord is 18 cm.

_____

The chord is tangent to the inner circle, so makes a 90° angle with the radius to that tangent point. This tells you that the triangle shown is a right triangle. It also tells you that the short radius bisects the chord. The Pythagorean theorem can be used to find the length of the side not shown (half the chord length).

The unknown side (a) can be found from ...

15² = 12² +a²

225 -144 = a² . . . . . . subtract 12²

81 = a² . . . . . . . . . . . simplify

9 = a . . . . . . . . . . . . . take the square root

The chord length is 2a, so is ...

2(9 cm) = 18 cm . . . . length of chord XY

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Given the diagram above, m∠Z is:

Mama L [17]

Answer:

z = 120º

Step-by-step explanation:

t + 2t + t + 2t = 360º

60 + 2(60) + 60 + 2(60) = 360

t = 60

2t = z

z = 120º

This is all based on the fact there there are 360º in a quadrilateral.

5 0

3 years ago

Help!!! Please!!!!!!

maw [93]

What are you supposed to do.

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2 years ago

A ball is thrown vertically upwards from the ground. It rises to a height of 10m and then falls and bounces. After each bounce,

Citrus2011 [14]

${\huge{\fcolorbox{yellow}{red}{\orange{\boxed{\boxed{\boxed{\boxed{\underbrace{\overbrace{\mathfrak{\pink{\fcolorbox{green}{blue}{Answer}}}}}}}}}}}}}$

(i)

$\sf{a_n = 20 \times {( \frac{2}{3} )}^{n - 1} }$

(ii)

$\sf S_n = 60 \{1 - { \frac{2}{3}}^{n} \}$

Step-by-step explanation:

$\underline\red{\textsf{Given :-}}$

height of ball (a) = 10m

fraction of height decreases by each bounce (r) = 2/3

$\underline\pink{\textsf{Solution :-}}$

(i) We will use here geometric progression formula to find height an times

${\blue{\sf{a_n = a {r}^{n - 1} }}} \\ \sf{a_n = 20 \times { \frac{2}{3} }^{n - 1} }$

(ii) here we will use the sum formula of geometric progression for finding the total nth impact

$\orange {\sf{S_n = a \times \frac{(1 - {r}^{n} )}{1 - r} }} \\ \sf S_n = 20 \times \frac{1 - ( { \frac{2}{3} })^{n} }{1 - \frac{2}{3} } \\ \sf S_n = 20 \times \frac{1 - {( \frac{2}{3}) }^{n} }{ \frac{1}{3} } \\ \sf S_n = 3 \times 20 \times \{1 - ( { \frac{2}{3}) }^{n} \} \\ \purple{\sf S_n = 60 \{1 - { \frac{2}{3} }^{n} \}}$