According to your text first equation is
y = 3x+2
second equation is
y = Ix-1I + 1
Definition of our absolute value Ix-1I is
1) we take x-1 with condition x-1≥0 => x≥1
2) we take -(x-1) = -x+1 with x<1
Let's form the first system with condition x≥1
y= 3x+2 and y=x-1+1=> y=x
If the left sides of the equations are equal, than the right sides are equal too
3x+2=x => 3x-x=-2 => 2x=-2 => x=-1
This solution we can't accept because it does not meet the condition
x≥1
Lets's form the second system with condition x<1
y=3x+2 and y= -x+1+1 =. y = -x+2
In the same way as previous the right sides are equal =>
3x+2=-x+2 => 3x+x = +2-2 => 4x=0 => x=0
We accept this solution bacause its meet with condition x<1
When we replace this solution in equation y=3x+2 we got
y=3*0+2=2 Only one solution is (x,y) = (0,2)
Good luck !!!!!
A 3x+3c
B 9x^2+12x
C -4ab+24a
D 3x-6
E 3x+3
F (x+1)(2x+3)
G x^2+x-6
H 6x^2-31x+18
I 6x-8y+11
J 2x^3+11x^2+17x+6
M=12
40
44
20
250/11
Last one is 33
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Answer:
No
Step-by-step explanation:
Because it is terminal
I think it’s 3, i input it all into my calculator. i’m not sure about the exponent tho so probably 3^29
