I think it is 0.05 but i do if it is right your welcome if it is wrong I'm sorry
My answer is 13+16+19 because we’re are adding 3 each time
Answer:
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Step-by-step explanation:
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Answer:
Step-by-step explanation:
Directions
- Draw a circle
- Dear a chord with a length of 24 inside the circle. You just have to label it as 24
- Draw a radius that is perpendicular and a bisector through the chord
- Draw a radius that is from the center of the circle to one end of the chord.
- Label where the perpendicular radius to the chord intersect. Call it E.
- You should get something that looks like the diagram below. The only thing you have to do is put in the point E which is the midpoint of CB.
Givens
AC = 13 inches Given
CB = 24 inches Given
CE = 12 inches Construction and property of a midpoint.
So what we have now is a right triangle (ACE) with the right angle at E.
What we seek is AE
Formula
AC^2 = CE^2 + AE^2
13^2 = 12^2 + AE^2
169 = 144 + AE^2 Subtract 144 from both sides.
169 - 144 = 144-144 + AE^2 Combine
25 = AE^2 Take the square root of both sides
√25 = √AE^2
5 = AE
Answer
The 24 inch chord is 5 inches from the center of the circle.
The only way 3 digits can have product 24 is
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4
To be divisible by 3 the sum of the digits must be divisible by 3.
1+ 3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3.
So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.
Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.
The others must end in 3.
They must be greater than 152 which is 225. So the
First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.
The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.