Answer:
Step-by-step explanation:
yes the value of x and y of the intersection point is the solution.
Answer:
B.
Step-by-step explanation:
5(y - 6) means 'Multiply the difference between y and 6 by 5'.
The shortest side is 130 feet, the longest side is 260 feet and the greatest possible area is 33800 square feet
<h3>What dimensions would guarantee that the garden has the greatest possible area?</h3>
The given parameter is
Perimeter, P = 520 feet
Represent the shorter side with x and the longer side with y
One side of the garden is bordered by a river:
So the perimeter is:
P = 2x + y
Substitute P = 520
2x + y = 520
Make y the subject
y = 520 - 2x
The area is
A = xy
Substitute y = 520 - 2x in A = xy
A = x(520 - 2x)
Expand
A = 520x - 2x^2
Differentiate
A' = 520 - 4x
Set to 0
520 - 4x = 0
Rewrite as:
4x= 520
Divide by 4
x= 130
Substitute x= 130 in y = 520 - 2x
y = 520 - 2 *130
Evaluate
y = 260
The area is then calculated as:
A = xy
This gives
A = 130 * 260
Evaluate
A = 33800
Hence, the shortest side is 130 feet, the longest side is 260 feet and the greatest possible area is 33800 square feet
Read more about area at:
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Answer:
A. 7,348
Step-by-step explanation:
P = le^kt
intitial population = 500
time = 4 hrs
end population = 3,000
So we have all these variables and we need to solve for what the end population will be if we change the time to 6 hours. First, we need to find the rate of the growth(k) so we can plug it back in. The given formula shows a exponencial growth formula. (A = Pe^rt) A is end amount, P is start amount, e is a constant that you can probably find on your graphing calculator, r is the rate, and t is time.
A = Pe^rt
3,000 = 500e^r4
now we can solve for r
divide both sides by 500
6 = e^r4
now because the variable is in the exponent, we have to use a log
ln(6) = 4r
we can plug the log into a calculator to get
1.79 = 4r
divide both sides by 4
r = .448
now lets plug it back in
A = 500e^(.448)(6 hrs)
A = 7351.12
This is closest to answer A. 7,348
I hope this helps you
distance=speed×time
first cars time t
second cars time t+3
same distance
distance=55.t
distance=40. (t+3)
55t=40 (t+3)
55t-40t=120
15t=120
t=8
t+3=11
