Y = 4x + 1....slope here is 4. A parallel line will have the same slope
y = mx + b
slope(m) = 4
(3,9)...x = 3 and y = 9
now we sub and find b, the y int
9 = 4(3) + b
9 = 12 + b
9 - 12 = b
-3 = b
so ur parallel equation is : y = 4x - 3
On the first day of school last year, Cody was 148.5 cm tall.
<span>Perimeter =2w+2L= 520.
We can solve this by understanding that the area is maximized by a square
Therefore L=w.
p=2w+2w=520=4w
w=130
Area
A=wL=130(130)= 16900 square yards</span>
We can find this using the formula: L= ∫√1+ (y')² dx
First we want to solve for y by taking the 1/2 power of both sides:
y=(4(x+1)³)^1/2
y=2(x+1)^3/2
Now, we can take the derivative using the chain rule:
y'=3(x+1)^1/2
We can then square this, so it can be plugged directly into the formula:
(y')²=(3√x+1)²
<span>(y')²=9(x+1)
</span>(y')²=9x+9
We can then plug this into the formula:
L= ∫√1+9x+9 dx *I can't type in the bounds directly on the integral, but the upper bound is 1 and the lower bound is 0
L= ∫(9x+10)^1/2 dx *use u-substitution to solve
L= ∫u^1/2 (du/9)
L= 1/9 ∫u^1/2 du
L= 1/9[(2/3)u^3/2]
L= 2/27 [(9x+10)^3/2] *upper bound is 1 and lower bound is 0
L= 2/27 [19^3/2-10^3/2]
L= 2/27 [√6859 - √1000]
L=3.792318765
The length of the curve is 2/27 [√6859 - √1000] or <span>3.792318765 </span>units.
