Answer:
<u>Option D: A cylinder with a circumference of about 50 units</u>
Step-by-step explanation:
The rest of the question is the attached figure.
The square shown has a perimeter of 32 units. The square is rotated about line k. What shape is created by the rotation and what is the approximate circumference of the base? Circumference of a circle: C = 2πr
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Given: the perimeter of the square = 32
perimeter of a square is four times the length of its side
So, the side length of square = 32/4 = 8 units
The square is rotated about line k.
So, it will form a cylinder with radius 8 units.
Circumference of a circle = 2 π r
Where, r is the radius of the circle.
Circumference of the base is C = 2 π * 8
∴ C = 16π = 16 * 3.14
∴ C = 50.26548 ≈ 50
The shape is created by the rotation is a cylinder with a circumference of about 50 units.
<u>So, the answer option is D.</u>
For this case we have the following quadratic equation:
The solutions will be given by:
Where:
Substituting the values we have:
We have two roots:
ANswer:
The keyword is Rise over run
So the 2nd y value minus the first y value and divide it by the 2nd x value minus
(y2-y1)/(x2-x1)
(10-7)/5-3)
the first x value and you should get 3/2
*dont worry the if the number is under and not above it’s not a saying to do to the power of but it’s just saying the order in which to put the x and y values.
The first one is 4 and the last one is 2.14237024 u can put all of that in if u want
Let x represent the first even integer: x = 2(k) → y = 2k
Let y represent the second even integer: y = 2(k + 1) → y = 2k + 2
x · y = 48
(2k) · (2k + 2) = 48
4k² + 4k - 48 = 0
4(k² + k - 12) = 0
4 (k + 3)(k - 2) = 0
k = -3, k = 2 (Note: POSITIVE integers so k = -3 is ruled out)
x = 2k → x = 2(2) → x = 4
y = 2(k + 1) → y = 2(2 + 1) → y = 2(3) → y = 6
The smaller number (x) is 4
