Answer:
6
Step-by-step explanation:
Short AnswerThere are two numbers
x1 = -0.25 + 0.9682i <<<<
answer 1x2 = - 0.25 - 0.9582i <<<<
answer 2 I take it there are two such numbers.
Let one number = x
Let one number = y
x + y = -0.5
y = - 0.5 - x (1)
xy = 1 (2)
Put equation 1 into equation 2
xy = 1
x(-0.5 - x) = 1
-0.5x - x^2 = 1 Subtract 1 from both sides.
-0.5x - x^2 - 1 = 0 Order these by powers
-x^2 - 0.5x -1 = 0 Multiply though by - 1
x^2 + 0.5x + 1 = 0 Use the quadratic formula to solve this.

a = 1
b = 0.5
c = 1

x = [-0.5 +/- sqrt(0.25 - 4)] / 2
x = [-0.5 +/- sqrt(-3.75)] / 2
x = [-0.25 +/- 0.9682i
x1 = -0.25 + 0.9682 i
x2 = -0.25 - 0.9682 i
These two are conjugates. They will add as x1 + x2 = -0.25 - 0.25 = - 0.50.
The complex parts cancel out. Getting them to multiply to 1 will be a little more difficult. I'll do that under the check.
Check(-0.25 - 0.9682i)(-0.25 + 0.9682i)
Use FOIL
F:-0.25 * -0.25 = 0.0625
O: -0.25*0.9682i
I: +0.25*0.9682i
L: -0.9682i*0.9682i = - 0.9375 i^2 = 0.9375
NoticeThe two middle terms (labled "O" and "I" ) cancel out. They are of opposite signs.
The final result is 0.9375 and 0.0625 add up to 1
Answer:
56
Step-by-step explanation:
Answer:f⁻¹(x) =
Explanation:To get the inverse of a function, we have to follow these steps:1- replace f(x) with y
2- swap the x and y variables
3- solve for y
4- replace y with f⁻¹(x)
Applying this to the given as follows:f(x) =
Step 1: replace f(x) with y:y =
Step 2: swap the x and y variables:x =
Step 3: solve for y:7x = 4y - 3
4y = 7x - 3
y =
Step 4: replace y with f⁻¹(x):f⁻¹(x) =

Hope this helps :)