The most exact answer would be 0.19951% but the approximate answer would be 0.2%.
Answer:
2x -y ≥ 4
Step-by-step explanation:
The intercepts of the boundary line are given, so it is convenient to start with the equation of that line in intercept form:
... x/(x-intercept) + y/(y-intercept) = 1
... x/2 + y/(-4) = 1
Multiplying by 4 gives the equation of the line.
... 2x -y = 4
This line divides the plane into two half-planes. The half-plane that is shaded is the one for larger values of x and/or smaller values of y than the ones on the line. So, for some given y, if we increase x we will get a number from our equation above that is greater than 4. Hence, the inequality we want is ...
... 2x -y ≥ 4
We use the ≥ symbol because the line is solid, so part of the solution space.
Answer:
Option (A)
Step-by-step explanation:
It has been given in this question that sign telling path has a 2% grade.
2% grade means a rise of 2 meters for a horizontal change of 100 m (As given in the figure attached).
All the trigonometric ratios for the angle θ between the path and the horizontal are,
Sinθ = 
Cosθ = 
tanθ = 
Since measures of the opposite side and adjacent sides are given
Therefore, tangent ratio will be applied to get the measure of the angle,
tanθ = 
θ = 
Option (A) will be the answer.
Answer:
Step-by-step explanation:
The volume of a rectanguiar shape like this one is V = L * W * H, where the letters represent Length, Width and Height. Here L is the longest dimension and is 28 - 2x; W is the width and is 22-2x; and finally, x is the height. Thus, the volume of this box must be
V(x) = (28 - 2x)*(22 - 2x)*x
and we want to maximize V(x).
One way of doing that is to graph V(x) and look for any local maximum of the graph. We'd want to determine the value of x for which V(x) is a maximum.
Another way, for those who know some calculus, is to use the first and second derivatives to identify the value of x at which V is at a maximum.
I have provided the function that you requested. If you'd like for us to go all the way to a solution, please repost your question.