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
Given:
The function is
To find:
The asymptotes and zero of the function.
Solution:
We have,
For zeroes, f(x)=0.
Therefore, zero of the function is 0.
For vertical asymptote equate the denominator of the function equal to 0.
Taking square root on both sides, we get
So, vertical asymptotes are x=-4 and x=4.
Since degree of denominator is greater than degree of numerator, therefore, the horizontal asymptote is y=0.
solution is (4, - 7 )
given the equations
y = 2x - 15 → (1)
4x + 3y = - 5 → (2)
substitute y = 2x - 15 into (2)
4x + 3(2x - 15) = - 5
4x + 6x - 45 = - 5
10x - 45 = - 5 ( add 45 to both sides )
10x = 40 ( divide both sides by 10 )
x = 4
substitute x = 4 into (1) for corresponding value of y
y = (2 × 4 ) - 15 = 8 - 15 = - 7
solution is (4, - 7 )
Answer:
3.5 cm
Step-by-step explanation:
In the picture attached, circle M is shown.
From the picture, minor arc JH measures 25°.
The length of the circle is π*Diameter = π*16 = 50.26 cm. This length is proportional to 360°. To find the length of minor arc JH, we have to use the next proportion:
50.26 cm / x cm = 360° / 25°
x = 50.26*25/360
x = 3.5 cm
