Answer:
f(g(x)) = 2(x^2 + 2x)^2
f(g(x)) = 2x^4 + 8x^3 + 8x^2
Step-by-step explanation:
Given;
f(x) = 2x^2
g(x) = x^2 + 2x
To derive the expression for f(g(x)), we will substitute x in f(x) with g(x).
f(g(x)) = 2(g(x))^2
f(g(x)) = 2(x^2 + 2x)^2
Expanding the equation;
f(g(x)) = 2(x^2 + 2x)(x^2 + 2x)
f(g(x)) = 2(x^4 + 2x^3 + 2x^3 + 4x^2)
f(g(x)) = 2(x^4 + 4x^3 + 4x^2)
f(g(x)) = 2x^4 + 8x^3 + 8x^2
Hope this helps...
Answer:
Step-by-step explanation:
Y=-2x+2.....(1)
Y=-2x-2.....(2)
Subtracting (1) from (2)
Y - Y = -2x - (-2x) -2 - 2
O = -2x + 2x - 4
O = 0 - 4
Hence it as no solution the the two variables tend to zero
Answer:
A. Right angle -- if you multiply the slopes of 2 intersecting lines and you "-1" then the lines are perpendicular - in this case the line in quadrant 3 has a slope of "1" and the line in quadrant 4 has a slope of "-1", hence their product is "-1" and the angles formed at the intersection of the lines are right angles.
Step-by-step explanation:
Answer:
A. No, the student is not right. The central limit theorem says nothing about the histogram of the sample values. It deals only with the distribution of the sample means.
Step-by-step explanation:
No, the student is not right. The central limit theorem says nothing about the histogram of the sample values. It deals only with the distribution of the sample means. The central limit theorem says that if we take a large sample (i.e., a sample of size n > 30) of any distribution with finite mean 
and standard deviation 
, then, the sample average is approximately normally distributed with mean and variance 
.
Perimeter = (2 × Length) + (2 × Width)
If the length is dependent on the width (because the length is 10 more than the width), we can say that L = W + 10.
To find the answer to the Peri, we have to find out what the W is first.
184 = 2(W + 10) + 2W
184 = 2W + 20 + 2W
184 = 4W + 20
184 - 20 = 4W = 164
164 ÷ 4 = W
41 = W
We have found out width. Our length is W + 10, so L = 41 + 10 = 51
W = 41 and L = 51
