What is the scale factor of the dilation of ABCD centered at the origin?
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Answer: Hours by Tina practiced this month is 3 times the hours she practiced last month such that 15 x 3 =45 hours.
Step-by-step explanation:
Given : Duration of Tina practiced piano last month= 15 hours
Duration of Tina practiced piano this month= 45 hours
We can see 45 is a multiple of 15 such that
Therefore, Hours by Tina practiced this month is 3 times the hours she practiced last month such that 15 x 3= 45 hours.
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![-20 \leq 3x+1 \leq 10\\ \Rightarrow\ -21 \leq 3x \leq 9\\ \Rightarrow\ -7 \leq x \leq 3\\ Sol=[-7\ 3]\ if\ x\ is\ a\ real. ](https://tex.z-dn.net/?f=-20%20%5Cleq%203x%2B1%20%5Cleq%2010%5C%5C%0A%5CRightarrow%5C%20-21%20%5Cleq%203x%20%5Cleq%209%5C%5C%0A%5CRightarrow%5C%20-7%20%5Cleq%20x%20%5Cleq%203%5C%5C%0A%0ASol%3D%5B-7%5C%203%5D%5C%20if%5C%20x%5C%20is%5C%20a%5C%20real.%0A)
1)
∠BAC = ∠NAC - ∠NAB = 144 - 68 = 76⁰
AB = 370 m
AC = 510 m
To find BC we can use cosine law.
a² = b² + c² -2bc*cos A
|BC|² = |AC|²+|AB|² - 2|AC|*|AB|*cos(∠BAC)
|BC|² = 510²+370² - 2*510*370*cos(∠76⁰) =
|BC| ≈ 553 m
2)
To find ∠ACB, we are going to use law of sine.
sin(∠BAC)/|BC| = sin(∠ACB)/|AB|
sin(76⁰)/553 m = sin(∠ACB)/370 m
sin(∠ACB)=(370*sin(76⁰))/553 =0.6492
∠ACB = 40.48⁰≈ 40⁰
3)
∠BAC = 76⁰
∠ACB = 40⁰
∠CBA = 180-(76+40) = 64⁰
Bearing C from B =360⁰- 64⁰-(180-68) = 184⁰
4)
Shortest distance from A to BC is height (h) from A to BC.
We know that area of the triangle
A= (1/2)|AB|*|AC|* sin(∠BAC) =(1/2)*370*510*sin(76⁰).
Also, area the same triangle
A= (1/2)|BC|*h = (1/2)*553*h.
So, we can write
(1/2)*370*510*sin(76⁰) =(1/2)*553*h
370*510*sin(76⁰) =553*h
h= 370*510*sin(76⁰) / 553= 331 m
h=331 m
