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When we have consecutive values, f(10), f(11), f(12), f(13), f(14), we can make a difference table to determine the degree of f as a polynomial. A quadratic will have a constant second difference:
x 10 11 12 13 14
f(x) 50 71 94 119 146
1st diff 21 23 25 27
2nd diff 2 2 2
We got a constant second difference, so f is a polynomial of degree two.
Answer: This function is quadratic
Answer:
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Step-by-step explanation:
<u>Solution-</u>
Given that,
In the parallelogram PQRS has PQ=RS=8 cm and diagonal QS= 10 cm.
Then considering ΔPQT and ΔSTF,
1- ∠FTS ≅ ∠PTQ ( ∵ These two are vertical angles)
2- ∠TFS ≅ ∠TPQ ( ∵ These two are alternate interior angles)
3- ∠TSF ≅ ∠TQP ( ∵ These two are also alternate interior angles)
<em>If the corresponding angles of two triangles are congruent, then they are said to be similar and the corresponding sides are in proportion.</em>
∴ ΔFTS ∼ ΔPTQ, so corresponding side lengths are in proportion.

As QS = TQ + TS = 10 (given)
If TS is x, then TQ will be 10-x. Then putting these values in the equation



∴ So TS = 3.85 cm and TQ is 10-3.85 = 6.15 cm