Answer:
Step-by-step explanation:
Let's FOIL this out and get it into standard quadratic format:
. The lack of a linear term in the middle means therewas no upwards velocity, consistent with the object being dropped straight down as opposed to thrown up in the air tand then falling in a parabolic path. The -4.9t² represents the acceleration due to gravity, and the 490 represents the height from which the object was dropped. The constant in a quadratic that is modeling parabolic motion always represents the height from which the object was dropped (or launched). That's how you know.
Answer:
R2 = 6.44
T2 = 8.82
Whole = 30.42
Step-by-step explanation:
R1 = 2.8 X 3.1
T1 = (1/2) X (2.4) X (5.4)
R2 = (5.4-3.1) X (2.8) = 6.44
T2 = (1/2) X (11.7-5.4) X (2.8) = 8.82
Whole = (1/2) X (11.7) X (2.4+2.8) = 30.42
a)
Check the picture below.
b)
volume wise, we know the smaller pyramid is 1/8 th of the whole pyramid, so the volume of the whole pyramid must be 8/8 th.
Now, if we take off 1/8 th of the volume of whole pyramid, what the whole pyramid is left with is 7/8 th of its total volume, and that 7/8 th is the truncated part, because the 1/8 we chopped off from it, is the volume of the tiny pyramid atop.
Now, what's the ratio of the tiny pyramid to the truncated bottom?
Answer:
x = 500 yd
y = 250 yd
A(max) = 125000 yd²
Step-by-step explanation:
Let´s call x the side parallel to the stream ( only one side to be fenced )
y the other side of the rectangular area
Then the perimeter of the rectangle is p = 2*x + 2* y ( but only 1 x will be fenced)
p = x + 2*y
1000 = x + 2 * y ⇒ y = (1000 - x )/ 2
And A(r) = x * y
Are as fuction of x
A(x) = x * ( 1000 - x ) / 2
A(x) = 1000*x / 2 - x² / 2
A´(x) = 500 - 2*x/2
A´(x) = 0 500 - x = 0
x = 500 yd
To find out if this value will bring function A to a maximum value we get the second derivative
C´´(x) = -1 C´´(x) < 0 then efectevly we got a maximum at x = 500
The side y = ( 1000 - x ) / 2
y = 500/ 2
y = 250 yd
A(max) = 250 * 500
A(max) = 125000 yd²
