First, write an equation system based on the problem
An equation for "the perimeter of a rectangle is 70 in" can be written as following
∴ 2l + 2w = 70 <em>(first equation)</em>
An equation for "the length is 11 in more than its width" can be written as following
∴ l = w + 11 <em>(second equation)</em>
Second, solve the equation by substitution method o find the dimension of the rectangle.
Substitute/plug l as (w+11) into the first equation
2l + 2w = 70
2(w + 11) + 2w = 70
2w + 22 + 2w = 70
4w + 22 = 70
4w = 70 - 22
4w = 48
w = 48/4
w = 12
The width of the rectangle is 12 in
Substitute w with 12 to the second equation
l = w + 11
l = 12 + 11
l = 23
The length of the rectangle is 23 in
Third, find the area of the rectangle
a = l × w
a = 23 × 12
a = 276
The area of the rectangle is 276 in²
I will try i had this unit last year, basically the diameter is 5 in this problem, which you multiply by pi (in this case I will use 3.14 as the shortened version since its on-going) which gives you 15.7, then you multiply that by 3 in this problem to get 47.1!
The vertex is the point where the graph is horizontal, which you can find by setting the first derivative to zero.
derivative: -4x + 8 = 0 => x=2
fill it in in the equation:
y=-2*2² + 8*2 - 5 = 3
So the answer is the point (2,3)
Use desmos graphing calculator hope that helps
