Answer:
Step-by-step explanation:
Represent the length of one side of the base be s and the height by h. Then the volume of the box is V = s^2*h; this is to be maximized.
The constraints are as follows: 2s + h = 114 in. Solving for h, we get 114 - 2s = h.
Substituting 114 - 2s for h in the volume formula, we obtain:
V = s^2*(114 - 2s), or V = 114s^2 - 2s^3, or V = 2*(s^2)(57 - s)
This is to be maximized. To accomplish this, find the first derivative of this formula for V, set the result equal to 0 and solve for s:
dV
----- = 2[(s^2)(-1) + (57 - s)(2s)] = 0 = 2s^2(-1) + 114s - 2s^2
ds
Simplifying this, we get dV/ds = -4s^2 + 114s = 0. Then either s = 28.5 or s = 0.
Then the area of the base is 28.5^2 in^2 and the height is 114 - 2(28.5) = 57 in
and the volume is V = s^2(h) = 46,298.25 in^3
Idk what linear combination is but you can solve this way:
3x+y=4
-2x-y=-5
x=-1
3(-1)+y=4
-3+y=4
y=7
So x=-1 and y=7
Answer: B. 2.5 in
Step-by-step explanation:
From the given right angle triangle,
the hypotenuse of the right angle triangle is the unknown side.
With m∠32 as the reference angle,
the adjacent side of the right angle triangle is 4 in
the opposite side of the right angle triangle is w
To determine w, we would apply
the tangent trigonometric ratio which is expressed as
Tan θ = opposite side/adjacent side. Therefore,
Tan 32 = w/4
w = 4tan32 = 4 × 0.625
w = 2.5 in
Answer:
y ≤ 1/3 x - 1.3.
Step-by-step explanation:
First find the equation of the line:
The slope = (-0.3 - (-1.3) / (3 - 0).
= 1/3 so the equation is y = 1/3x + b where b is a constant.
b is the y-intercept which is the value of y when x = 0 . We see that it is -1.3 ( from the point 0, -1.3).
Since the line is continuous and the shading is below this line the inequality sign is 'less than or equal to', ≤.
