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Zepler [3.9K]
3 years ago
6

AB and AD are tangent to circle C. Find the length of AB. A. 39 B. 16 C. 3 D. 13

Mathematics
1 answer:
MrMuchimi3 years ago
7 0

Answer:

B. 16

Step-by-step explanation:

Since AB and AD are tangent to circle C, they equal each other.

2x + 7 = 3x - 9     Make them equal each other

2x + 7 = 3x - 9      Subtract 2x from both sides

7 = x - 9            Add 9 to both sides

16 =  x            

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The last step in a proof contains the
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Answer:

Conclusion

Step-by-step explanation:

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3 years ago
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Can someone solve these please using the formula AB | a - b |<br><br> And can you show work
Lana71 [14]

Answer: 1) AC = 13

The formua does not actually apply to all of the problems.

Step-by-step explanation:

1) The absolute value of -8 is added to the absolute value of 5. 8+5=13

2) Subtract the length of the EF from EG to get the length of FG 21 -6 = 15

3) Take what's given and create an equation to solve. 4x + 15 +39 =110. 4x = 110 - (15+39).

4x =110-54. 4x =56. x=56/4. x=14

4) Create another equation. You have two segments that add up to the length of EG, given =23

EF+FG=EG

(2x-12)+(3x-15)=23

5x - 27 = 23

5x= 23+27 5x =50. x = 10

Substitute 10 for x

EF=2(10) -12 EF=8

FG=3(10)-15. FG=15

EF+FG =EG.

8 + 15 = 23

5) 2/5 of 25 is 10 So EF is 10. Subtract from 25 to get FG

FG = 15

I hope this helps you.

4 0
3 years ago
What is 12/24 - 10/23
Elodia [21]
The LCD of 24 and 23 is 552, so 276/552 - 240/552 = 36. After simplifying 36/552, you get 3/46 and that is your answer.
4 0
2 years ago
There are 5 radar stations and the probability of a single radar station detecting an enemy plane is 0.55. What is the probabili
Lerok [7]
This is a binomial distribution with n = 5, p = 0.55, q = 1 - 0.55 = 0.45, x = 0, 1, 2, 3

P(x) = nCx p^x q^(n - x)
P(x ≤ 3) = 1 - P(x > 3) = 1 - [P(4) + P(5)]
P(4) = 5C4 x (0.55)^4 x (0.45) = 0.2059
P(5) = 5C5 x (0.55)^5 x 1 = 0.0503

P(x ≤ 3) = 1 - (0.2059 + 0.0503) = 1 - 0.2562 = 0.7438
4 0
3 years ago
Simplify
OlgaM077 [116]

Answer:

see below

Step-by-step explanation:

\frac{x-1}{x^2-3x+2}+ \frac{x-2}{x^2-5x+6} +\frac{x-5}{x^2-8x+15}

we need to simplify that

x^2-3x+2=(x-1)(x-2)\\\\x^2-5x+6=(x-2)(x-3)\\\\x^2-8x+15=(x-3)(x-5)

so we can continue

\frac{x-1}{(x-1)(x-2)}=\frac{1}{x-2}\\\\\frac{x-2}{(x-2)(x-3)} =\frac{1}{x-3}\\\\\frac{x-5}{(x-3)(x-5)} =\frac{1}{x-3}

and we can put all together

\frac{1}{x-2}+ \frac{1}{x-3}+ \frac{1}{x-3}\\\\\frac{1}{x-2} +\frac{2}{x-3}\\\\\frac{x-3}{(x-3)(x-2)}+ \frac{2(x-2)}{(x-2)(x-3)} \\\\\frac{x-3+2x-4}{(x-3)(x-2)}\\\\\frac{3x-7}{x^2-5x+6}

3 0
3 years ago
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