Looks correct to me but i'm not 100% sure
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
Answer:
a. (15, 15)
Step-by-step explanation:
We start with those two equations:
1) a - 1.2b = -3
2) 0.2b + 0.6a = 12
We'll begin by modifying equation #1 to isolate a:
a = -3 + 1.2b
Then we'll use this value for a in the second equation:
0.2b + 0.6 (-3 + 1.2b) = 12
0.2b - 1.8 + 0.72b = 12
0.92b = 13.8
b = 15
Then we'll place that value of b in the first equation to find a:
a - 1.2 (15) = -3
a - 18 = -3
a = 15
Answer:
n?
Step-by-step explanation:
Answer:
80
Step-by-step explanation:
3/5 of 200 = 120
200-120 = 80