C(x) = 300.
Our question becomes simple: 300 = x^2-40x+610
x^2-40x+610-300 = 0
x^2-40x+310 = 0
Now find the value for x with either the quadratic formula or by factorization.
Let me know how it goes. Giving the full answer won’t help you improve
SA=2H(L+W)+LW for open box
20=2H(4+W)+4(W)
distribute
20=8H+2WH+4W
divide both sides by 2
10=4H+WH+2W
solve for 1 variable, pick W
10-4H=WH+2W
10-4H=W(H+2)
(10-4H)/(H+2)=W
V=LWH
subsitute 4 for V, subsitute H for H and (10-4H)/(H+2) for W
V=(4)(H)(10-4H)/(H+2)
V=(40H-16H²)/(H+2)
find max value
take deritivitive of this thing
V'=-16(H²+4H-5)/((H+2)²)
using sign chart
sign changes from positive to negative at H=1
so at H=1
find W
W=(10-4H)/(H+2)
W=2
the dimeionts are
length=4ft
width=2ft
height=1ft
(the volume is 8 cubic feet)
Answer:
<h2>3.6°</h2>
Step-by-step explanation:
The question is incomplete. Here is the complete question.
Find the angle between the given vectors to the nearest tenth of a degree.
u = <8, 7>, v = <9, 7>
we will be using the formula below to calculate the angle between the two vectors;

is the angle between the two vectors.
u = 8i + 7j and v = 9i+7j
u*v = (8i + 7j )*(9i + 7j )
u*v = 8(9) + 7(7)
u*v = 72+49
u*v = 121
|u| = √8²+7²
|u| = √64+49
|u| = √113
|v| = √9²+7²
|v| = √81+49
|v| = √130
Substituting the values into the formula;
121= √113*√130 cos θ
cos θ = 121/121.20
cos θ = 0.998
θ = cos⁻¹0.998
θ = 3.6° (to nearest tenth)
Hence, the angle between the given vectors is 3.6°