How does tripling the side lengths of a right triangle affects its area?
1 answer:
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The function f be described by 
The graph of this function, is the graph of the parent function
, which is the parabola with vertex at (0, 0) which opes upwards:
i) reflected with respect to the y-axis, because of the minus
ii) shifted 100 units up
check the picture attached.
A function has an inverse only if it is decreasing or increasing.
So we must divide the following cases for which the inverses exist:
i) with domain (-infinity, 0]
ii) with domain [0, infinity)
To find the formula for the inverse function g of f, we use the property:
f(g(x))=x
thus,
![f(g(x))=x\\\\ 100-[g(x)]^2=x\\\\g^2(x)=100-x\\\\g(x)= \mp\sqrt{100-x}](https://tex.z-dn.net/?f=f%28g%28x%29%29%3Dx%5C%5C%5C%5C%20100-%5Bg%28x%29%5D%5E2%3Dx%5C%5C%5C%5Cg%5E2%28x%29%3D100-x%5C%5C%5C%5Cg%28x%29%3D%20%5Cmp%5Csqrt%7B100-x%7D%20)
so each of - and + cases, are the inverses of
i) with domain (-infinity, 0]
ii) with domain [0, infinity)
At this point, recall that the domain of a function, is the range of it's inverse function,
and the range of the function, is the domain of the inverse function.
thus,
the range of
, which is [0,infinity) is the domain of
i) with domain [0,infinity)
So our answer is:
inverse of i) with domain (-infinity, 0] is 
and the inverse of ii) with domain [0, infinity) is
The graph of this function, is the graph of the parent function
i) reflected with respect to the y-axis, because of the minus
ii) shifted 100 units up
check the picture attached.
A function has an inverse only if it is decreasing or increasing.
So we must divide the following cases for which the inverses exist:
i)
ii)
To find the formula for the inverse function g of f, we use the property:
f(g(x))=x
thus,
so each of - and + cases, are the inverses of
i)
ii)
At this point, recall that the domain of a function, is the range of it's inverse function,
and the range of the function, is the domain of the inverse function.
thus,
the range of
i)
So our answer is:
inverse of i)
and the inverse of ii)
The ship is 88.8 miles far from the port at 4 pm.
Given,
The displacement from 1 to 3 in the afternoon.
D1 = 30 miles/hr × 2 hr
= 60 miles to the north
The displacement from 3 till 4 in the afternoon.The ship changes its course 20 degrees eastward at 3 o'clock.
Therefore, D2 = 30 miles/hr × 1 hr
= 30 miles to the 20° north eastward
By combining two vectors, the resulting displacement:
D = √((D1 ₊ D2 × cos(20°))² ₊ (D2 × sin(20°))²
D = √((60 ₊ 30 × cos (20°))² ₊ (30 × sin(20°))²
D = 88.8 miles
Hence the ship is 88.8 miles far away from the port at 4 pm.
Learn more about "Law of sines and cosines" here-
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