Area= 190
height= 2(width) - 1
area= length times width
190= (2(width) -1) x width
190 = 2w^2-w
w^2-w-190 = 0
use quadratic fomula, you'll get w= 10
so height will be 2(10)-1 = 19
This is a problem of permutation or arrangements, the word PRECALCULUS has 11 letters, of which 3 letters appear twice, we want to get the number of indistinguishable arrangements, this is how we do it,
![P_r^n=\frac{n!}{(n-r)!}](https://tex.z-dn.net/?f=P_r%5En%3D%5Cfrac%7Bn%21%7D%7B%28n-r%29%21%7D)
For our problem, what we do is
Answer:
2 ×
- 1
Step-by-step explanation:
Answer:
y = 70 yd
A(max) = 9800 yd²
Step-by-step explanation:
Let call " y " sides of the rectangular area running perpendicular to the river, then:
Perimeter of the area (without one side is )
P = x + 2*y and we have 280 yards of fencing material
280 = x + 2*y ⇒ y = ( 280 - x ) / 2
And rectangular area is:
A = x*y
Area as a function of x is:
A(x) = x * ( 280 - x ) / 2
A(x) = 280*x /2 - x²/2
Taking derivatives on both sides of the equation we get:
A´(x) = 140 - x = 0
140 - x = 0
x = 140 yd
And y
y = ( 280 - x ) / 2
y = 140 /2
y = 70 yd
And
A( max ) = 140*70
A(max) = 9800 yd²
The answer is y=3x-6!!!!!