<u>We are Given:</u><u>_______________________________________________</u>
ΔABC right angled at B
BC = 8
AC = 20
<u>Part A:</u><u>_____________________________________________________</u>
Finding the length of AB
From the Pythagoras theorem, we know that:
AC² = BC² + AB²
replacing the given values
(20)² = (8)² + AB²
400 = 64 + AB²
AB² = 336 [subtracting 64 from both sides]
AB = 18.3 [taking the square root of both sides]
<u>Part B:</u><u>_____________________________________________________</u>
Finding Sin(A)
we know that Sin(θ) = Opposite / Hypotenuse
The side opposite to ∠A is BC and The hypotenuse is AC
So, Sin(A) = BC / AC
Sin(A) = 8/20 [plugging the values]
Sin(A) = 2/5
<u>Part C:</u><u>_____________________________________________________</u>
Finding Cos(A)
We know that Cos(θ) = Adjacent / Hypotenuse
The Side adjacent to ∠A is AB and the hypotenuse is AC
So, Cos(A) = AB / AC
Cos(A) = 18.3/20 [plugging the values]
Cos(A) = 183 / 200
<u>Part D:</u><u>_____________________________________________________</u>
Finding Tan(A)
We know that Tan(θ) = Opposite / Adjacent
Since BC is opposite and AB is adjacent to ∠A
Tan(A) = BC / AB
Tan(A) = 8 / 18.3 [plugging the values]
Tan(A) = 80 / 183
Answer:
Step-by-step explanation:
for this graph find rise over run , and where the line crosses they y axis.. sooo ..
it crosses at -2 and the slope is rise over run or y/x and I can see by the graph that the line goes up 1 when it goes over 1... so the slope is +1
sooo...
use the slope intercept form y = mx + b
y = 1x + (-2)
y = x - 2 :)
Answer is 10.5 miles
I used Pythagorean theory to solve for the unknown distance from the waterfall to lookout point.
a^2 + b^2 = c^2
2^2 + 4^2 = c^2
20 = c^2
Take square root of both sides
4.47 = c
Rounded to 4.5 miles
Add 4 miles and 2 miles and 4.5 miles
10.5 miles total
Answer:
n = 4
Step-by-step explanation:
![\sqrt[n]{256} = 8^{1/3} * 8^{1/3}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7B256%7D%20%3D%208%5E%7B1%2F3%7D%20%20%2A%208%5E%7B1%2F3%7D)
![\sqrt[n]{256} = 8^{2/3}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7B256%7D%20%3D%208%5E%7B2%2F3%7D)
![\sqrt[n]{2^8} = 8^{2/3}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7B2%5E8%7D%20%3D%208%5E%7B2%2F3%7D)




< --- This is your answer
Hope this helps!
Answer:
probably the second one or third one
Step-by-step explanation:
pretty sure! :)