Answer:
432 yd squared
Step-by-step explanation:
This is a parallelogram. The area of a parallelogram is denoted by: A = bh, where b is the base and h is the height. Here, the base is b = 19 1/5 and the height is h = 22 1/2. Plug these in:
A = bh = (19 1/5) * (22 1/2)
To make this simpler, let's convert the numbers into decimals. 1/5 is just 0.2 so 19 1/5 is 19.2. 1/2 is just 0.5, so 22 1/2 is 22.5. Now we have:
A = 19.2 * 22.5 = 432
Thus the area is 432 yd squared.
Hope this helps!
By definition, the volume of the cone is given by:
V = (1/3) (π) (r ^ 2) (h)
Substituting values we have:
V = (1/3) (π) ((4/2) ^ 2) (5)
V = (20/3) (π)
Then, the volume of the cylinder is:
Vc = 3 * V
Vc = 3 * (20/3) (π)
Vc = 20π
Answer:
the volume of the original cylinder jesse bought is:
Vc = 20π
A vertical line can only pass through the x axsis not the y unless it was at 0 then it would pass through all the y points. A diagonal line can pass through (-5,1) but not a vertical choose x=-5 if you want it to be a vertical line going through (-5,0)
Answer:
14
Step-by-step explanation:
the four smaller rectangles of each of the corners: 2*1=2. since there are four of them, 2*4=8
the middle, larger rectangle: (6-2-2)*(1+1+1) = 2*3 =6
add small and big rectangles: 8+6 = 14
Answer:
Subtract from both sides of the equation the term you don't want
Step-by-step explanation:
In solving equations, you generally want to "undo" operations that are done to the variable. Addition is "undone" by adding the opposite (that is, subtracting the amount that was added). Multiplication is "undone" by division.
If you have variables on both sides of the equation, pick one of the variable terms and subtract it from both sides of the equation.
<u>Example</u>
2x = x +1
If we choose to subtract x, then we will have a variable term on the left and a constant term on the right:
2x -x = x -x +1 . . . . . . . x is subtracted from both sides
x = 1 . . . . . . simplify
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Note that we purposely set up this example so that removing the variable term from the right side caused the variable term and constant term to be on opposite sides of the equal sign. It may not always be that way. As long as you remember that an unwanted term can be removed by subtracting it (from both sides of the equation), you can deal with constant terms and variable terms no matter where they appear.
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<em>Additional Comment</em>
It usually works well to choose the variable term with the smallest (or most negative) coefficient. That way, when you subtract it, you will be left with a variable term that has a positive coefficient.
