X = E/W dimension
<span>y = N/S dimension </span>
<span>4x + 4x + 2y + 2y = 64 </span>
<span>8x + 4y = 64 </span>
<span>4y = 64 - 8x </span>
<span>y = 16 - 2x </span>
<span>Area = xy = x(16 - 2x) = 16x - 2x^2 </span>
<span>Maximum of y = ax^2 + bx + c is when x = -b / 2a </span>
<span>so x = -16 / -4 = 4 </span>
Answer:
coffee is the best answer in my mind
We have that the spring is going to have a sin or a cos equation. We have that the maximum distance of the spring is 6 inches and it is achieved at t=0. Let's fix this as the positive edge. Until now, we have that the function is of the form:
6sin(at+B). We have that the period is 4 minutes and hence that the time component in the equation needs to make a period (2pi) in 4 minutes. Thus 4min*a=2p, a=2p/4=pi/2. In general, a=2pi/T where a is this coefficient, T is the period. Finally, for B, since sin(pi/2)=1, we have that B=pi/2 because when t=0, we have that 6sin(B)=6. Substituting, we have f(t)=6sin(pi*t/2+pi/2)=6cos(pi*t/2)
by trigonometric identities.
Alice should pick the enlarged-photo with dimensions of 8-inch by 10-inch.
Step-by-step explanation:
Step 1:
In order for a part of the photo to not be cut off, the enlarged photo's dimensions should be of a constant ratio with the original photo's dimensions.
We divide the dimensions of the enlarged-photo with the dimensions of the original photo to check which has a constant ratio.
Step 2:
The original photo was a 4-inch by 5-inch photo.
Option 1 is 7-inch by 9-inch, so the ratios are
The ratios are different so this cannot be the enlarged photo's dimensions.
Option 2 is 8-inch by 10-inch, so the ratios are
The ratios are the same so this can be the enlarged photo's dimensions.
Option 3 is 12-inch by 16-inch, so the ratios are
The ratios are different so this cannot be the enlarged photo's dimensions.
So the enlarged-photo with dimensions of 8-inch by 10-inch should be picked.
