As the input is increased by 1.
4 / (1/4) = 4 * 4 = 16
64/4 = 16
1024/64 = 16
So we can see the output factor is increased by a factor of 16.
Hope this helps.
Answer:
Solution : y = ± 1/3x
Step-by-step explanation:
Remember that the equation of a hyperbola is in the form (y - k)² / a² - (x - h)² / b² = 1. Therefore we have a² here = 9, and b² here = 81.
To determine the asymptotes we can use the equation y = k ± a/b(x - h). What makes this equation really simple, is that k = 0 and h = 0, giving us y = ± a/b(x). Let's isolate a and b given a² = 9, and b² = 81 --- (1)
a² = 9
a = 3
b² = 81
b = 9
Therefore our asymptotes will be in the form y = ± a/b(x) = ± 3/9(x) = ± 1/3x. Our solution is thus option a.
-3x^4 + 27x^2 + 1200 = 0
factor out -3
-3(x^4 -9x^2 -400)=0
divide by -3
x^4 -9x^2 -400=0
(x^2-25) (x^+16) = 0
x^2-25=0 x^2+16=0
(x-5)(x+5)=0 x^2=-16
x=5 x=-5 x = 4i x=-4i
Answer: x=5 x=-5 x = 4i x=-4i
Answer:
To get maximum area, the dimensions will be;
l = 200 and b = 100
Step-by-step explanation:
Given the data in the question;
No fencing is needed on this side as the farmer wants to fence 3 sides of a rectangular field with 400 m fencing.
now, let the two vertical length sides be x meter and the horizontal length be 400 - 2x
as shown in the image below;
so,
Area of rectangle = (400 - 2x) × x = 400x - 2x²
now, to maximize area 'A', we will make A' = 0
⇒ 400 - 2(2x) = 0
400 = 4x
x = 400/4 = 100
so
⇒ (400 - 2x) = 400 - 2(100) = 400 - 200 = 200
∴ Maximum area = ( 400 - 2x) × x
= 200 × 100
= 20000 m²
∴ To get maximum area, the dimensions will be;
l = 200 and b = 100
If shapes have at least one pair of right angles, they do have one pair of perpendicular sides (True).
