Find four consecutive integers whose sum is 114.
1 answer:
27, 28, 29 and 30
consecutive integers have a difference of 1 between each one
let n be the first integer then the next three are
n + 1, n + 2 and n + 3
hence n + n + 1 + n + 2 + n + 3 = 114
4n + 6 = 114
subtract 6 from both sides of the equation
4n = 114 - 6 = 108
divide both sides by 4
n = = 27
n = 27, n+ 1 = 28, n + 2 = 29, n + 3 = 30
the consecutive integers are 27, 28, 29 and 30
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Answer:
Option (C)
Step-by-step explanation:
From the graph attached,
In ΔABC and ΔA'B'C',
∠A ≅ ∠A'
∠B ≅ ∠B'
Therefore, ΔABC ~ ΔA'B'C'
Corresponding sides of these triangles will be proportional.
Therefore, ratio of the sides, AC : A'C' = 1 : 3 shows that image triangle A'B'C' is a dilated form of pre-image ABC with a scale factor of 3.
Option (C) will be the correct option.