Answer:
250 smaller boxes
Step-by-step explanation:
Find the volume of the 2 cuboids.
Formula = Length x width x height
Big cuboid = 50 x 30 x 20 = 30000
Small cuboid = 10 x 3 x 4 = 120
Now divide them.
30000 ÷ 120 = 250 small boxes
Answer:
−
6
x + 3
Step-by-step explanation:
hope that helps :)
Answer:
Mean wage is the average wage for a worker in a specified position or occupation, which may be skewed up or down if there are a few extreme examples in the sample.
Step-by-step explanation:
Answer:
The value of x that maximizes the volume enclosed by this box is 0.46 inches
The maximum volume is 3.02 cubic inches
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
The volume of the open-topped box is equal to
where
substitute
Convert to expanded form
using a graphing tool
Graph the cubic equation
Remember that
The domain for x is the interval -----> (0,1)
Because
If x>1
then
the width is negative (W=2-2x)
so
The maximum is the point (0.46,3.02)
see the attached figure
therefore
The value of x that maximizes the volume enclosed by this box is 0.46 inches
The maximum volume is 3.02 cubic inches
Answer:
g(-4) = -1
g(-1) = -1
g(1) = 3
Explanation:
If you are given a function that is defined by a system of equations associated with certain intervals of x, just find which interval makes x true, and then substitute x into the equation of that interval.
For example, given g(-4), this is an expression which is asking for the value of the equation when x = -4. So -4 is not ≥ 2, so ¼x - 1 will not be used. -4 is also not ≤ -1 and ≤ 2, so -(x - 1)² + 3 will not be used either. So in turn, we will just use -1 which is always -1 so g(-4) will just be -1, right because there is no x variable in -1 so it will always be the same.
Using the same idea as before g(-1) is g(x) when x = -1 so -1 will not be a solution because -1 is not less than -1 (< -1). -1 is not ≥ 2 either so we will be using the second equation because -1 is part of the interval -1≤x≤2 (it is a solution to this inequality), therefore -(x - 1)² + 3 will be used.
As x = -1, -(x - 1)² + 3 = -(-1 - 1)² + 3 = -(-2)² + 3 = -4 + 3 = -1.
It is a coincidence that g(-1) = -1.
Now for g(1), where g(x) has an input of 1 or the value of the function where x = 1, we will not use the first equation because x = 1 → x < -1 → 1 < -1 [this is false because 1 is never less than -1], so we will not use -1.
We will use -(x - 1)² + 3 again because 1 is not ≥ 2, 1≥2 [this is also false]. And -1 ≤ 1 < 2 [This is a true statement]. Therefore g(1) = -(1 - 1)² + 3 = -(0)² + 3 = 3
