The heat produced by current I is
H = I²R
where
R = resistance.
According to the formula, heat produced is proportional to the square of the current.
When a current of I = 2 amps is applied, the heat produced is
H = 10x³ + 80.
This heat includes heat due to a fixed current of 2 amps, and heat due to a variable current of x amps.
Because the heat produced is proportional to the square of the current, write the expressions as
H = (10x)*(x²) + 20*(2²)
The second term on the right is heat due to the fixed current of 2 amps, written as
20*(2²).
Therefore the fixed resistance is R = 20 ohms, and the square of the fixed current is 2².
The first term represents heat due to variable resistance, written as
(10x)*(x²).
Therefore the variable resistance is 10x, and the square of the variable current is x².
Answer:
The variable current is 10x.
Answer:
0?
Step-by-step explanation:
Answer:
Step-by-step explanation:
Range goes from the lowest point on a graph to the highest point, not including anything in between those 2 extremes. We will evaluate the quadratic for the values given. First f(2):
so
f(2) = -1
Now, f(4):
so
f(4) = 11
Now, f(7):
so
f(7) = 44
The lowest y value is -1 and the highest is 44, so the range is from -1 to 44 (I can't tell which of your choices reflects that because it's too small!)
Answer:
Students:Adults = 118:94
Step-by-step explanation:
212-118=94
Answer:
Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, ![[ab]_K](https://tex.z-dn.net/?f=%20%5Bab%5D_K%20)
is equal to ![[ba]_K](https://tex.z-dn.net/?f=%20%5Bba%5D_K%20)
, thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.
Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, ![[ab]_H = [ba]_H](https://tex.z-dn.net/?f=%20%5Bab%5D_H%20%3D%20%5Bba%5D_H%20)
. Since a and b were generic elements of H, then H/G is abelian.
