B.addition property of multiplication
D.inverse property of multiplication
E.commutative property of addition
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work
EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same
For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y
For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4
Therefore, for the two expressions to be conjugates, we must satisfy the two conditions.
Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the
x²y = -4 ... (I)
Condition 2: Real parts are the same
x² + y = -3 ... (II)
We have a system of equations since both conditions must be satisfied
x²y = -4 ... (I)
x² + y = -3 ... (II)
We can rearrange equation (II) so that we have
y = -3 - x² ... (II)
Substituting into equation (I)
x²y = -4 ... (I)
x²(-3 - x²) = -4
-3x² - x⁴ = -4
x⁴ + 3x² - 4 = 0
(x² + 4)(x² - 1) = 0
(x² + 4)(x-1)(x+1) = 0
Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.
Solve for y:
y = -3 - x² ... (II)
y = -3 - (±1)²
y = -3 - 1
y = -4
So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:
-3 + ix²y
= -3 + i(±1)²(-4)
= -3 - 4i
x² + y + 4i
= (±1)² - 4 + 4i
= 1 - 4 + 4i
= -3 + 4i
They result in conjugates
Find what is common between 35 and 55....and that would be 5....so factor 5 out
35x + 55 =
5(7x + 11) <==
Answer:
The function f(x)= –StartRoot negative x EndRoot is shown on the graph.
On a coordinate plane, an absolute value graph starts at (0, 0) and goes down and to the left through (negative 4, negative 2).
Step-by-step explanation:
The range of the graph is all real numbers greater than or equal to 0.
Answer:
x=-2 (the actual solution)
x=-5 (the extraneous solution)
Step-by-step explanation:
First step is to isolate the radical part (the part with the square root).
So we will add 4 on both sides first:

To get rid of square we are going to square both sides:

We are going to expand right hand side (you can use "F.O.I.L":

Subtract x on both sides and subtract 6 on both sides:

Combine like terms:

We will now factor:
(Since 2(5)=10 and 2+5=7)
Set both factors equal to 0:
x+2=0 or x+5=0
Solve first equation by subtracting 2 on both sides to obtain x=-2.
Solve second equation by subtracting 5 on both sides to obtain x=-5.
Now let's check to see if these are both actually indeed solutions to the equation we began with:
for
:



is a true equation so
is indeed a solution.
for
:



is false so
is what we call the extraneous solution.