Answer:
The area of the triangles are equal.
E, the rest are all linear functions
Answer:
15839
Step-by-step explanation:
14000 x 1.025^5
<h3><em>In AP form 2nd term - 1st term = 3rd term - 2nd term
</em></h3><h3><em>b²-a² = c²-b²
</em></h3><h3><em>b²+b² = c²+a²
</em></h3><h3><em>2b² = c²+a²
</em></h3><h3><em>
</em></h3><h3><em>Add 2ab+2ac+2bc on both sides
</em></h3><h3><em>
</em></h3><h3><em>2b²+2ab+2ac+2bc = a²+c²+ac+ac+bc+bc+ab+ab
</em></h3><h3><em>2b²+2ab+2ac+2bc = ac+bc+a²+ab+bc+c²+ab+ac
</em></h3><h3><em>2b²+2ab+2ca+2cb = ca+cb+a²+ab+cb+c²+ab+ac
</em></h3><h3><em>2(ba+b²+ca+cb) = (ca+cb+a²+ab) + (cb+c²+ab+ac)
</em></h3><h3><em>2((ba+b²)+(ca+cb)) = ((ca+cb)+(a²+ab)) + ((cb+c²)+(ab+ac))
</em></h3><h3><em>2(b(a+b)+c(a+b)) = (c(a+b)+a(a+b)) + (c(b+c)+a(b+c)) </em></h3><h3><em>2(b+c)(a+b) = (c+a)(a+b) + (c+a)(b+c)
</em></h3><h3><em>
</em></h3><h3><em>Divide whole by (a+b)(b+c)(c+a)</em></h3><h3><em></em></h3><h3><em>2/c+a = 1/b+c + 1/a+b</em></h3><h3><em>1/c+a + 1/c+a = 1/b+c + 1/a+b</em></h3><h3><em>1/c+a - 1/b+c = 1/a+b - 1/c+a</em></h3><h3><em></em></h3><h3><em>2nd term - 1st term = 3rd term - 2nd term
</em></h3><h3><em>Thus 1/b+c, 1/c+a, 1/a+b are in AP.</em></h3><h3><em></em></h3><h3><em>HOPE IT HELPS !!!</em></h3><h3><em>THANK YOU !!!</em></h3>
Check the picture below.
make sure your calculator is in Degree mode.
