5 consecutive even integers: 2n, 2n+2, 2n+4, 2n+6, 2n+8 for any integer n The sum of the two smallest of five consecutive even integers is 50 less than the sum of the other three integers: 2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 504n + 2 = 6n -3234 = 2nn = 17 <span>The smallest integer in our sequence is 2n = 2*17 = 34</span> Check:2n + 2n+2 = 2n+4 + 2n+6 + 2n+8 - 5034 + 36 = 38 + 40 + 42 - 5070 = 120 - 5070 = 70
Answer:
The solutions are x = 1.24 and x = -3.24
Step-by-step explanation:
Hi there!
First, let´s write the equation:
log[(x² + 2x -3)⁴] = 0
Apply the logarithm property: log(xᵃ) = a log(x)
4 log[(x² + 2x -3)⁴] = 0
Divide by 4 both sides
log(x² + 2x -3) = 0
if log(x² + 2x -3) = 0, then x² + 2x -3 = 1 because only log 1 = 0
x² + 2x -3 = 1
Subtract 1 at both sides of the equation
x² + 2x -4 = 0
Using the quadratic formula let´s solve this quadratic equation:
a = 1
b = 2
c = -4
x = [-b± √(b² - 4ac)]/2a
x = [-2 + √(4 - 4(-4)·1)]/2 = 1.24
and
x = [-2 - √(4 - 4(-4)·1)]/2 = -3.24
The solutions are x = 1.24 and x = -3.24
Have a nice day!
The system of equations that represent the matrix is c
If you divide all of them together starting inside of the parentheses you will get the wrong answer bc I’m just typing random thing
