Answer:
volume= 56.55 yds
Step-by-step explanation:
If you're looking for the volume, the formula is 
.
Diameter is two times the radius (radius is half of the diameter) so the radius of the cylinder would be 1.5 yds.
After you've gotten that information, plug those values into the formula.
Solve 1.5^2 first. Then multiply that value by pi. Then multiply that value by the height which is 8yds.
Volume=56.55yds rounded to the nearest hunderedth.
<span>y=x(x+5)(x-8)
The zeros of </span><span>y=x(x+5)(x-8), means to solve the value of x, when y =0
</span><span>y=x(x+5)(x-8) = 0
</span><span>
x(x+5)(x-8) = 0
x = 0 or (x+5) = 0 or (x - 8) = 0
x = 0 x + 5 = 0 x - 8 = 0
x = 0 -5 x = 0 + 8
x = -5 x = 8
Hence the zeros are x = 0, x = -5, x = 8</span>
Solution,
y = -x-4......(i)
3x + y =- 16.......(ii)
Now,
Substituting the value of y in equation (ii)
or, 3x + (-x-4)= -16
or,3x - x - 4 = -16
or, 2x= -16 +4
or, x = -12/2
: x = -6
Then,
Substituting the value of x in (i) equation;
or,y= -(-6) - 4
or,y= 6 - 4
: y =2
To start, let x represent the width and x+100 represent the length. Since the perimeter of a figure is the sum of all the measurements of the side which can be represented by (x+100)+(x+100)+x+x and since you know your perimeter is 1220, you can set the expression equal to 1220. This would look like this:
(x+100)+(x+100)+x+x=1220
Once you have done that, combine any like terms (combine terms with the same variables and raised to the same power together) which would simplify to this:
4x+200=1220
Now that you have your like terms simplified, subtract 200 from both sides to get 4x=1020 and finally, to solve for x, or find the width, divide both sides by 4 to get x=255.
Now that you have your width, now you must find your length as the question asks to find the dimensions of the rectangular field. To find the length, add 100 to the width, 255 since according to the information given, the length is 100 more than the width. When you add 100 to 255, you should get that your length is 355.
Now that you have your length and width, you can conclude that the dimensions of the field is 255 by 355 feet, which is your answer :)
