Answer:
The answer is 3.1
If we were to convert it into a **FRACTION** the answer would be : 31/10.
And that i an improper fraction, but as a **MIXED NUMBER** :
All answers would be : 3.1 , 31/10 and 3 1/10
Find the perimeter of the figure to the nearest hundredth. perimeter: about cm
1 answer:
The perimeter of the figure to the nearest hundredth is 50.26 cm.
<h3>Perimeter of a figure.</h3>The perimeter of a figure is the sum of the whole sides of the figure.
The perimeter of the figure can be found as follows:
perimeter of the figure = 12 + 5 + 5 + circumference of the semi circle
circumference of the semi circle = πr
where
- r = radius
perimeter of the figure = 22 + πr
perimeter of the figure = 22 + 3.14(9)
perimeter of the figure = 22 + 28.26
perimeter of the figure = 50.26 cm
learn more on perimeter here: brainly.com/question/11419989
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Answer:
x = 4, y = 1
(4, 1)
Step-by-step explanation:
-2x - y = -9
5x -2y = 18
-2x - y = -9
+2x +2x
-y = 2x - 9
/-1 /-1
y = -2x + 9
5x -2y = 18
5x - 2(-2x + 9) = 18
5x + 4x - 18 = 18
9x - 18 = 18
+ 18 + 18
9x = 36
/9 /9
x = 4
Now, we solve for y:
-2(4) - y = -9
-8 - y = -9
+ 8 +8
-y = -1
/-1 /-1
y = 1
(x, y) -> (4, 1)
Hope this helps!
For a given girth, the largest area will be that of a square. Then for some length x, the volume will be
.. V = x((108 -x)/4)^2
.. = (1/16)*x*(108 -x)^2
The largest volume will be found where the derivative of this with respect to x is zero.
.. V' = (1/16)*((108 -x)^2 -2x(108 -x))
.. 0 = (108 -x)(108 -x -2x)
.. x = 108 or 108/3 = 36
The dimensions of the largest package that can be mailed are
.. 36 inches in length and 72 inches in girth
The volume of that package is 11,664 in^3.
.. V = x((108 -x)/4)^2
.. = (1/16)*x*(108 -x)^2
The largest volume will be found where the derivative of this with respect to x is zero.
.. V' = (1/16)*((108 -x)^2 -2x(108 -x))
.. 0 = (108 -x)(108 -x -2x)
.. x = 108 or 108/3 = 36
The dimensions of the largest package that can be mailed are
.. 36 inches in length and 72 inches in girth
The volume of that package is 11,664 in^3.