The formula for illuminance is given by
E = I / d^2
This formula only holds true for one-dimensional illuminance
The problem asks for the illuminance across the floor. We need to use two variables, x and y.
From Pythagorean Theorem
d^2 = x^2 + y^2
and from Trigonometry
x = d cos t
y = d sin t
The function for the illuminance can be represented by the composite function
E = I cos² t / x²
and
E = I sin² t / y²
The boundary of these functions is:
<span>0 < t < 8
So, the value of t must be in radians and not in degrees</span>
Answer:
∠6=58°
Step-by-step explanation:
∠3=∠6=58°
It would be: 63 / 28 = 9 / 4
So, your answer is 9/4
Hope it helped.
Answer:
(16x + 21) and (16x - 6)
Step-by-step explanation:
f(g(x)) = f(6 + 4x)
Applying the f(x) function on (6 + 4x) gives
4(6 + 4x) - 3
Which equals 16x + 24 - 3
= 16x + 21
g(f(x)) = g(4x - 3)
Applying the g(x) function on (4x - 3) gives
6 + 4(4x - 3)
Which equals 6 + 16x - 12
= 16x - 6
Answer:
The answer to your question is the first option
Step-by-step explanation:
64
Process
1.- Find the prime factors of 64
64 2
32 2
16 2
8 2
4 2
2 2
1
64 = 2⁶
2.- Express 64 as a fractional exponent
64
3.- Simplify
64
64
64![^{1/4} = 2\sqrt[4]{2^{2}}](https://tex.z-dn.net/?f=%5E%7B1%2F4%7D%20%3D%202%5Csqrt%5B4%5D%7B2%5E%7B2%7D%7D)
4.- Result
64![^{1/4} = 2\sqrt[4]{4}](https://tex.z-dn.net/?f=%5E%7B1%2F4%7D%20%3D%202%5Csqrt%5B4%5D%7B4%7D)
