The perimeter of each is the sum of the lengths of its sides.
Triangle Perimeter = 16 + 4x-3 +3x = 7x+13
Square Perimeter = 2x +2x +2x +2x = 8x
If these are equal, you have
The length of a side of the square is 26 units.
The answer would most likely be C
Suppose the length of the triangle is x, if the perimeter of the rectangle is 100 ft, the width of the rectangle will be (50-x) ft.
Area of rectangle will be:
A=length*width
A=x(50-x)
A=50x-x^2
at maximum area, dA/dx=0
thus
dA/dx=50-2x=0
solving for x we get
2x=50
x=25
thus for maximum area length=25 ft
the size of the width will be
50-x=50-25=25 ft
thus the maximum area will be:
25*25=625 sq. feet
Solving #19
<u>Take y-values from the graph</u>
- a) (g·f)(-1) = g(f(-1)) = g(1) = 4
- b) (g·f)(6) = g(f(6)) = g(2) = 2
- c) (f·g)(6) = f(g(6)) = f(5) = 1
- d) (f·g)(4) = f(g(4)) = f(2) = -2
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.
