Answer:
A) Dimensions;
Length = 20 m and width = 10 m
B) A_max = 200 m²
Step-by-step explanation:
Let x and y represent width and length respectively.
He has 40 metres to use and he wants to enclose 3 sides.
Thus;
2x + y = 40 - - - - (eq 1)
Area of a rectangle = length x width
Thus;
A = xy - - - (eq 2)
From equation 1;
Y = 40 - 2x
Plugging this for y in eq 2;
A = x(40 - 2x)
A = 40x - 2x²
The parabola opens downwards and so the x-value of the maximum point is;
x = -b/2a
Thus;
x = -40/2(-2)
x = 10 m
Put 10 for x in eq 1 to get;
2(10) + y = 40
20 + y = 40
y = 40 - 20
y = 20m
Thus, maximum area is;
A_max = 10 × 20
A_max = 200 m²
X-the number
20 · x² = 20 · x |divide both sides by 20
x² = x |subtract x from both sides
x² - x = 0
x(x - 1) = 0 ⇔ x =0 or x - 1 = 0
Answer: x = 0 or x = 1
Answer:
55 m²
Step-by-step explanation:
The arc length of a half-circle is πr, so the area of the curved surface is ...
A = πrl = π(2.5 m)(7 m) = 17.5π m² ≈ 55 m²
Y-y1=m(x-x1)
m= slope =-3
y-y1=-3(x-x1),
y1=-7, x1=5
y--7=-3(x-5)
y+7=-3(x-5) this is C
