So 12 mins to 600 feet
35 mins=1900 feet
find the slope which is change in top over change in bottom (aka slope) so to find do
(y1-y2)/(x1-x2)
we want, per minute, so put the time as x1 and x2
x1=12
y1=600
x2=35
y2=1900
subsitutte
(600-1900)/(12-35)=-1300/-23=56.5 per minute
the answe ris 56.5 feet per minute or the last answer
Answer:
<u>B. 7(x − 5)(y + 2)</u>
Explanation:
A. 7(2x − 5)(y + 2) = 14xy + 28x − 35y − 70 (Wrong)
<u><em>B. 7(x − 5)(y + 2) = 7xy + 14x − 35y − 70 (Correct)</em></u>
C. 7(x − 2)(y + 5) = 7xy <u>+</u> 35x− 14y − 70 (Wrong)
D. 7(x − 10)(y + 2) = 7xy + 14x − 70y − 140 (Wrong)
Answer:
200 √17
Step-by-step explanation:
distance between two points (x,y) and (x',y')
D² = (x' - x)² + (y' - y)²
D² = (500 - (-300))² + (400 - 200)² = 640000 + 40000 = 680000
D = √680000 = √4 x 17 x 10000 = 200 √17
Basically, what this asks you is to maximize the are A=ab where a and b are the sides of the recatangular area (b is the long side opposite to the river, a is the short side that also is the common fence of both corrals). Your maximization is constrained by the length of the fence, so you have to maximize subject to 3a+b=450 (drawing a sketch helps - again, b is the longer side opposite to the river, a are the three smaller parts restricting the corrals)
3a+b = 450
b = 450 - 3a
so the maximization max(ab) becomes
max(a(450-3a)=max(450a-3a^2)
Since this is in one variable, we can just take the derivative and set it equal to zero:
450-6a=0
6a=450
a=75
Plugging back into b=450-3a yields
b=450-3*75
b=450-225
b=215
Hope that helps!
