The expression is a perfect square trinomial if and only if c = 121.
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How to get the value of c?</h3>
A perfect square trinomial is written as:
(a + b)^2 = a^2 + b^2 + 2ab
In this case, we have:
t^2 - 22t + c
We can rewrite this as:
t^2 - 2*11*t + c
Then we have:
a = t
b = -11
And we will have that:
c = b^2 = (-11)^2 = 121
Then we have:
t^2 - 2*11*t + 121
Which is a perfect square trinomial:
(t - 11)^2 = t^2 - 22t + 121
Learn more about perfect squares:
brainly.com/question/1538726
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Answer:
C words
Step-by-step explanation:
Answer:
AB = 21 and DE = 23
Step-by-step explanation:
Given 2 intersecting chords inside the circle then
The products of the measures of the parts of one chord is equal to the products of the measures of the parts of the other chord, that is
x(x + 13) = (x + 10)(x + 1) ← distribute parenthesis on both sides
x² + 13x = x² + 11x + 10 ← subtract x² + 11x from both sides
2x = 10 ( divide both sides by 2 )
x = 5
Hence
AB = x + 10 + x + 1 = 2x + 11 = (2 × 5) + 11 = 10 + 11 = 21
DE = x + x + 13 = 2x + 13 = (2 × 5) + 13 = 10 + 13 = 23
