Let
x-----------> first <span>odd integer
x+2--------> second consecutive odd integer
x+4-------> third consecutive odd integer
we know that
(x+4)</span>²=15+x²+(x+2)²-------> x²+8x+16=15+x²+x²+4x+4
x²+8x+16=19+2x²+4x-------> x²-4x+3
x²-4x+3=0
using a graph tool----------> to calculate the quadratic equation
see the attached figure
the solution is
x=1
x=3
the answer is
the first odd integer x is 1
the second consecutive odd integer x+2 is 3
the third consecutive odd integer x+4 is 5
The correct answer is option C.
The solution is shown below:
(8 - 3i)(6 + 5i)
= 8(6 +5i) - 3i(6 + 5i)
= 48 + 40i - 18i - 15i²
= 22i + 48 - 15(-1)
= 22i + 48 + 15
= 22i + 63
= 63 + 22i
No sé, prueba 2,3,5,7,11,13....
Hay muchos números primos!
Answer:
Cos b
Step-by-step explanation:
1/2[sin(a+b)+sin(a-b)]
1/2[sin a cos b +cos a sin b + sin A cos B - cos A sin B]
1/2[2sin a cos b]
sin a cos b
Answer: B
Step-by-step explanation:
A = 0/360 ×π×r^2
=135/360×π×11×11
=142.55 yd^2