◆ COMPLEX NUMBERS ◆
125 ( cos 288 + i sin 288 ) can be written as -
125.e^i( 288)
125.e^i( 288 +360 )
125.e^i( 288+ 720)
[ As , multiples of 360 can be added to an angle without changing any trigonometric functions or sign ]
To find the cube root , take the cube root of above 3 expressions ,
We get -
5 e^( i 96 )
5 e^( i 216 )
5 e^( i 336 )
Now using Euler's formula , We rewrite above as -
5 ( cos 96 + i sin 96 )
5(c os 216 + i sin 216 )
5 ( cos 336 + i sin 336 ) Ans.
Answer:
It's not a real solution
Step-by-step explanation:
So we have the system of equations:
equation (1)
equation (2)
To use substitution, we are going to solve for one variable in one of our equations, and then we are going to replace that value in the other equation:
Solving for
in equation (2):
equation (3)
Replacing equation (3) in equation (1):
equation (4)
Replacing equation (4) in equation (3):
We can conclude that the solution of our system of equations is <span>
(7/5, 21/10)</span>
The answer is i thinks x=3
Answer:
The first set of consecutive even integers equals (8 , 6)
The second set is ( - 8 and - 6) which also works.
Step-by-step explanation:
Equation
(x)^2 + (x + 2)^2 = (x)(x + 2) + 52 Remove the brackets on both sides
Solution
x^2 + x^2 + 4x + 4 = x^2 + 2x + 52 Collect the like terms on the left
2x^2+ 4x+ 4 = x^2 + 2x + 52 Subtract right side from left
2x^2 - x^2 + 4x - 2x + 4 - 52 = 0 Collect the like terms
x^2 + 2x - 48 = 0 Factor
(x + 8)(x - 6) = 0
Answer
Try the one you know works.
x - 6 = 0
x = 6
Therefore the two integers are 6 and 8
6^2 + 8^2 = 100
6*8 + 52 = 100
So 6 and 8 is one set of consecutive even numbers that works.
========================
What about the other set.
x + 8 = 0
x = - 8
x and x + 2
- 8 and -8 + 2 = - 8, - 6
(- 8 )^2 + (- 6)^2 = 100
(-8)(-6) + 52 = 100
Both sets of consecutive numbers work.
