Answer:
I'll setup the problem and you can do the calculations
Step-by-step explanation:
a.

V = volume; D = diameter; H = height of can
b.
Area of the Tin is the total area of the 2 Tin circles
one circle's area A is

A = area; D = diameter
calculate A
Then multiply by 2
c.
The area of the cardboard is the circumstance of the circle
times the height

d.
add answers from b and c
Well for a rectangle, the area
A=Length x Width,
1764<span>=Length x Width,
Since </span><span>the width of the field is 13 m more than the length, the length will be equal to the with + 13,
L=W+13,
</span>1764<span>=LW
</span>1764=(W+13)(W)
W=36 meters
They want the perimeters, so add them up
2W+2L
=2(36)+2(49)
=170 meters
Answer:
B. y=1/2x-5
Step-by-step explanation:
From the graph, we read the slope of line JK.
slope = rise/run = -3/-6 = 1/2
The equation we need has the same slope.
y = mx + b
y = (1/2)x + b
It passes through point P(6, -2). Now we find b.
-2 = (1/2)(6) + b
-2 = 3 + b
b = -5
The equation is
y = (1/2)x - 5
since it has a diameter of 28, then its radius must be half that or 14.
![\textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=14 \end{cases}\implies A=\pi (14)^2\implies A=196\pi ~\hfill \stackrel{\stackrel{semi-circle}{half~that}}{98\pi }](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D14%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%20%2814%29%5E2%5Cimplies%20A%3D196%5Cpi%20~%5Chfill%20%5Cstackrel%7B%5Cstackrel%7Bsemi-circle%7D%7Bhalf~that%7D%7D%7B98%5Cpi%20%7D)