Triangle JKL has vertices J(−2, 2) , K(−3, −4) , and L(1, −2) .
Rule: (x, y)→(x + 8, y + 1 )
J’ (-2, 2) → (-2 + 8, 2 + 1 ) → (6, 3 )
K’ (-3, -4) → (-3 + 8, -4 + 1 ) → (5, -3 )
L’ (1, -2) → (1 + 8, -2 + 1 ) → (9, -1)
J’ (6,3)
K’ (5,-3)
L’ (9,-1)
Hope this helps!
Answer:
Irregular polygons are polygons that do not have equal sides or equal angles. When finding the area of an irregular polygon, partition the polygon into smaller regular polygon shapes. You can then plug those measurements into their respective area formulas and multiply to find the area of one part of the shape.
This is 16cm.
To solve this question, you would need to find the common denominator among the 2 fractions. In this case it would be
72.
Then multiply numbers to both of the fractions to get to the bottom number in both fractions equaling 72.
40/72 + 27/72 =67/72.
Try to reduce to lowest terms by dividing a common number from both of them if possible.
Answer: I think your answer is B, but I'm not positive
Step-by-step explanation:
[7×(-3)] × (-2)²
= (-21) × 4
= -84
