Find three consecutive positive integers such that the square of the first increased by twice the second is three less than four
1 answer:
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Answer:
Step-by-step explanation:
Convert the mixed number into an improper fraction:
Now convert fractions so that their denominators are the same:
As the denominators are now the same, we can simply add the numerators:
Now convert back into a mixed number (if required):
Answer:
i. ΔAXC ~ ΔCXB
ii. ΔBCX Is-congruent-to ΔACX
Step-by-step explanation:
From the given ΔABC, CX is the altitude of ΔABC; and also an angle bisector of <ACB.
So that:
m<AXC = m<BXC (right angle property)
m<ACX = m<BCX (congruent property)
m<ACX + m<AXC + m<CAX = (sum of angles in a triangle)
m<BCX + m<BXC + m<CBX = (sum of angles in a triangle)
Therefore, from the figure it can be deduced that;
i. ΔAXC ~ ΔCXB (Angle-Angle-Side, AAS, property)
ii. ΔBCX Is-congruent-to ΔACX (Angle-Angle-Side, AAS, property)