Let W = width of package
Let H = height of package
Let L = length of package
The perimeter cab be one of the following:
P = 2(L + W), or
P = 2(L + H)
The perimeter of the cross section cannot exceed 108 in.
When the width is 10 in, then
2(L + 10) <= 108
L + 10 <= 54
L <= 44 in
When the height is 15 in, then
2(L + 15) <= 108
L + 15 <= 54
L <= 39 in
To satisfy both of these conditions requires that L <= 39 in.
Answer: 39 inches
Answer:
12
Step-by-step explanation:
the shapes are similar which means the sides are proportional, you can set up a ratio using the sides as 10/40 = 3/x
solving that gives you x =12
Answer:
25
Step-by-step explanation:
We require to solve for n, hence
n(n + 1) = 325
multiply both sides by 2 to eliminate the fraction
n(n + 1) = 650
n² + n = 650
subtract 650 from both sides to have equation in standard form
n² + n - 650 = 0 ← in standard form
(n + 26)(n - 25) = 0 ← in factored form
equate each factor to zero and solve for n
n + 26 = 0 ⇒ n = - 26
n - 25 = 0 ⇒ n = 25
however, n > 0 ⇒ n = 25
Answer:
(C)x=11.6, y=23.2
Step-by-step explanation:
Using Theorem of Intersecting Secant and Tangent
Applying this theorem in the diagram, we have:


Next, we apply Theorem of Intersecting Chords
PV X VQ=SV X VR
4 X x= 2 X y
Recall earlier we got: x=11.6
2y=4 X 11.6
2y=46.4
Divide both sides by 2
y=46.4/2=23.2
Therefore: x=11.6, y=23.2
Answer:
Thus we find that velocity vector at time t is
(5t+15, 5t^2/2, 4t^2)
Step-by-step explanation:
given that acceleration vector is a funciton of time and at time t

v(t) can be obtained by integrating a(t)
v(t) = 
Thus we use the fact that acceleration is derivative of velocity and velocity is antiderivative of acceleration.
The arbitary constant normally used for integration C is here C vector = initial velocity (u0,v0,w0)
Position vector can be obtained by integrating v(t)
Thus we find that velocity vector at time t is
(5t+15, 5t^2/2, 4t^2)