Let each side perpendicular to the river be "x".Then the side parallel to the river is "400-2x".
The formula for Area = width*length
A(x) = x(400-2x)
A(x) = 400x - 2x^2
You have a quadratic with a = -2 and b = 400
Maximum Area occurs where x = -b/2a = -400/(2*-2) = 100 ft. (width)
length = 400-2x = 400-2*100 = 200 ft (length)
Answer:
The true statement is h < k ⇒ (3)
Step-by-step explanation:
∵ The equation of the graph is y = + k
∵ (-h, k) are the coordinates of its vertex point
∵ -h is the x-coordinate of the vertex point
∵ x-coordinate of the vertex point is -3
∴ -h = -3
→ Divide both sides by -1
∴ h = 3
∵ k is the y-coordinate of the vertex point
∵ y-coordinate of the vertex point is 4
∴ k = 4
∵ 3 < 4
→ That means h is smaller than k
∴ h < k
∴ The true statement is h < k
<span>In a triangle the sum of two sides is always greater than the third side.
9-9 < x < 9+9
0 < x < 18 </span>← answer
Based on the weight and the model that is given, it should be noted that W(t) in radians will be W(t) = 0.9cos(2πt/366) + 8.2.
<h3>
How to calculate the radian.</h3>
From the information, W(t) = a cos(bt) + d. Firstly, calculate the phase shift, b. At t= 0, the dog is at maximum weight, so the cosine function is also at a maximum. The cosine function is not shifted, so b = 1.
Then calculate d. The dog's average weight is 8.2 kg, so the mid-line d = 8.2. W(t) = a cos t + 8.2. Then calculate a, the dog's maximum weight is 9.1 kg. The deviation from the average is 9.1 kg - 8.2 kg = 0.9 kg. W(t) = 0.9cost + 8.2
Lastly, calculate t. The period p = 2π/b = 2π/1 = 2π. The conversion factor is 1 da =2π/365 rad. Therefore, the function with t in radians is W(t) = 0.9cos(2πt/365) + 8.2.
Learn more about radians on:
brainly.com/question/12939121
Answer:
Coincidental
Step-by-step explanation:
<em>If the two lines have the same two points, they must also have the same slope </em><em>and</em><em> the same y intercept. From this we can deduct that they are actually the same line. Two lines that are directly on top of each other are considered coincidental lines.</em>