Do recall that squaring and the *radical sign* cancel each other out... like so:(
)
= a
When you put it that way, it isn't enough :P
(
)
= a
(
)
=?
so you start with
(
)
=
8x+1=25 <-- subtract 1 to both sides
8x=24 <- divide 8 to both sides
x= 3
To find out if it's an extraneous solution ask yourself: It mustn't result in a radical that I like to call... 'illegal'. Plug it into the radicand 8x+1 and make sure you get something that is not a negative number.... so, DO you get a negative number when you plug in x = 3 into the radicand?
(extraneous solution is a invalid solution)
x=3 not extraneous
They are functions because they have the variables
the asnwer will be 7 cuz -7/-1
The equation for a circle is
(x-h)² + (y-k)² = r²
h = your given x
y = your given k
Let's plug everything in!
(x - 5) ² + (y - (-1)) ² = 12²
Your final equation is
(x - 5) ² + (y + 1) ² = 144
<em>Hope I helped! Comment or message me if you have any questions :) </em>
Answer:
C. Solving this equation results in the statement –48 = 48. Because this is a false statement, the equation has no solution.