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LenaWriter [7]
3 years ago
15

11. Use the above figure to answer the question. How long is arc ACB if the radius of the circle is 6? Round your answer to two

decimal places. A. 25.13 B. 18.86 C. 6.29 D. 12.57

Mathematics
2 answers:
d1i1m1o1n [39]3 years ago
7 0

Answer:25.13

Step-by-step explanation:


Varvara68 [4.7K]3 years ago
3 0
I have to assume that figures OAT and OBT are congruent.

If that is true, the center angle O (of the sector AOB= is 120° and the angle of the sector ACB is 360° - 120° = 240°

Now you can state that the circumferece of the circle / 360 ° = arc ACB / 240° and solve for arc ACB.

arc ACB = circumference * 240 / 360 = 2πr *240/360 = 2π(6)*240/360 = 25.13

Answer: option A. 25.13
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rewona [7]
The correct answer is the second option (b)
4 0
3 years ago
a triangle with vertices at A (-1,1), B(-2,1) C(-1,4) is translated. the image of vertex A has coordinate at (3,-1) determine th
Drupady [299]
For this case we have the following vertices:
 A (-1,1) --------> A '(3, -1)
 B (-2.1)
 C (-1,4)
 We note that the translation rule is:
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 Applying the translacion rule we have:
 B (-2, 1) -----------> (-2 + 4, 1-2) ------> B '(2, -1)
 C (-1, 4) -----------> (-1 + 4, 4-2) ------> C '(3, 2)
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7 0
3 years ago
Let P(n) be the statement that n! < nn where n is an integer greater than 1.
Ira Lisetskai [31]

Answer: See the step by step explanation.

Step-by-step explanation:

a) First, Let P(n) be the statement that n! < n^n

where n ≥ 2 is an integer (This is because we want the statement of P(2).

In this case the statement would be (n = 2): P(2) = 2! < 2^2

b) Now to prove this, let's complet the basis step:

We know that 2! = 2 * 1 = 2

and 2^2 = 2 * 2 = 4

Therefore: 2 < 4

c)  For this part, we'll say that the inductive hypothesis would be assuming that k! < k^k for some k ≥ 1

d) In this part, the only thing we need to know or prove is to show that P(k+1) is also true, given the inductive hypothesis in part c.

e) To prove that P(k+1) is true, let's solve the inductive hypothesis of k! < k^k:

(k + 1)! = (k + 1)k!  

(k + 1)k!  < (k + 1)^k  < (k + 1)(k + 1)^k

Since k < k+1 we have:

= (k + 1)^k+1

f) Finally, as the base and inductive steps are completed, the inequality is true for any integer for any n ≥ 1. If we had shown P(4)

as our basis step, then the inequality would only be proven for n ≥ 4.

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3 years ago
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Answer:

a b d

Step-by-step explanation:

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