Given:
plot of land doubles in size.
original plot area 200 by 300 meters
Area = 200m * 300m
A = 60,000 m² original plot area
60,000m² x 2 = 120,000 m²
(200m + x)(300m + x) = 120,000 m²
60,000m² + 200m*x + 300m * x + x² = 120,000 m²
x² + 500x + 60,000 = 120,000
x² + 500x + 60,000 - 120,000 = 0
x² + 500x - 60,000 = 0
x = -b <u>+</u> (√b² - 4ac)/2a
a = 1 ; b = 500 ; c = -60,000
x = -500 + (√500² - 4(1)(-60,000)) / 2(1)
x = (-500 + (√250,000 + 240,000)) / 2
x = (-500 + 700)/ 2
x = 200 / 2
x = 100 meters
(200 + x)(300 + x) = 120,000
(200 + 100)(300 + 100) = 120,000
300 * 400 = 120,000
120,000 = 120,000
I think the answer would be 2 because if you divide 4 into 2 it equals 2
Answer:
60
Step-by-step explanation:
We can use the pythagorean theorem to find the leg.
10^2+x=13^2
x=12
We now find the area given the height of the triangle.
10 X 12=120
120/2=60
y = 9ln(x)
<span>y' = 9x^-1 =9/x</span>
y'' = -9x^-2 =-9/x^2
curvature k = |y''| / (1 + (y')^2)^(3/2)
<span>= |-9/x^2| / (1 + (9/x)^2)^(3/2)
= (9/x^2) / (1 + 81/x^2)^(3/2)
= (9/x^2) / [(1/x^3) (x^2 + 81)^(3/2)]
= 9x(x^2 + 81)^(-3/2).
To maximize the curvature, </span>
we find where k' = 0. <span>
k' = 9 * (x^2 + 81)^(-3/2) + 9x * -3x(x^2 + 81)^(-5/2)
...= 9(x^2 + 81)^(-5/2) [(x^2 + 81) - 3x^2]
...= 9(81 - 2x^2)/(x^2 + 81)^(5/2)
Setting k' = 0 yields x = ±9/√2.
Since k' < 0 for x < -9/√2 and k' > 0 for x >
-9/√2 (and less than 9/√2),
we have a minimum at x = -9/√2.
Since k' > 0 for x < 9/√2 (and greater than 9/√2) and
k' < 0 for x > 9/√2,
we have a maximum at x = 9/√2. </span>
x=9/√2=6.36
<span>y=9 ln(x)=9ln(6.36)=16.66</span>
the
answer is
(x,y)=(6.36,16.66)
<span>The rocket that will reach a greater distance above the ground at its maximum height is y = -4x</span>² + 80x. The answer is letter B. This is because of the coefficient presented in the x² term. The number -4 is much greater than -6. Graphing these equations into a graphing paper or excel, you would get a parabola.
