Answer:
C.) 1 5/b boxes
Number of boxes left over
= 3 - 1/2 - 2/3
Convert each fraction to one in terms of the LCM:
= 18/6 - 3/6 - 4/6
= 11/6
= 1 5/6 boxes.
F (x) = x^2 - 2x + 1
f (x) = (-2)^2 - 2(-2) + 1
= 4 + 4 + 1
= 9
f (x) = ( 0 )^2 - 2 (0) + 1
= 0 - 0 + 1
= 1
0 = x^2 - 2x + 1
x^2 = 2x - 1 = 1
f (2) = 2^2 - 2(2) + 1
= 4 - 4 + 1
= 1
f (3) = 3^2 - 2(3) + 1
= 9 - 6 + 1
= 3 + 1
= 4
Answer:
The perimeter (to the nearest integer) is 9.
Step-by-step explanation:
The upper half of this figure is a triangle with height 3 and base 6. If we divide this vertically we get two congruent triangles of height 3 and base 3. Using the Pythagorean Theorem we find the length of the diagonal of one of these small triangles: (diagonal)^2 = 3^2 + 3^2, or (diagonal)^2 = 2*3^2.
Therefore the diagonal length is (diagonal) = 3√2, and thus the total length of the uppermost two sides of this figure is 6√2.
The lower half of the figure has the shape of a trapezoid. Its base is 4. Both to the left and to the right of the vertical centerline of this trapezoid is a triangle of base 1 and height 3; we need to find the length of the diagonal of one such triangle. Using the Pythagorean Theorem, we get
(diagonal)^2 = 1^2 + 3^2, or 1 + 9, or 10. Thus, the length of each diagonal is √10, and so two diagonals comes to 2√10.
Then the perimeter consists of the sum 2√10 + 4 + 6√2.
which, when done on a calculator, comes to 9.48. We must round this off to the nearest whole number, obtaining the final result 9.
Answer:
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