Given that B is the midpoint of line AC and line BC is congruent to line DE.
The following statements and reasons, proves that line AB is congruent to line DE.
Statement Reasons
1. B is the midpoint of line AC Given
2. Line AB is congruent to line BC. Midpoint of a line segment
3. Line BC is congruent to line DE Given
4. Line AB is congruent to line DE Transitive property
A² + b² = c²
3² + 4² = c²
9 + 16 = c²
C² = 25
C = 5
so
SM = 5
In the first 10 -> 10 - 9 = 1 (contain a 3).
In the first 100 -> 100 - 9 * 9 = 19 (contain a 3).
In the first 10^n ->
10^n - 9^n (contain a 3).
<u>The answer is 3.439 numbers contain a 3 in the first 10.000</u>
<u>Answer</u>
12 units
<u>Explanation</u>
This can be solved using the properties of a circle.
DG and EG are secants f the circle.
DG×HG =EG×FG
(x+3+5)×5 = (x+6)×6
(x+8)5 = 6x + 36
5x +40 = 6x + 36
x = 4
DG = (x+3+5)
= 4 + 3 + 5
= 12 units