3x - 1 = 3x + 1
subtract 3x from both sides to get (-1 = 1). This is a false statement so it is: CONTRADICTION
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4x - 11 = 7
add 11 to both sides and then divide both sides by 4 to get 
. This statement is true only when x = 
so it is: CONDITIONAL
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2 - 8x = 2 - 8x
add 8x to both sides to get (2 = 2). This is a true statement so it is: IDENTITY
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x + 1 = -x + 4
add x to both sides, subtract 1 from both sides, and divide both sides by 2 to get 
. This statement is true only when x = 
so it is: CONDITIONAL
Answer: B, A, C, A
Well you can just keep adding 55 =1hr until you get to 400 the same with 45
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➷ It would be done in the same way it would if it was an integer ;
think of it as two separate parts
(3/5)^3 = 3/5 * 3/5 * 3/5
Calculate the top half, then the bottom half and put them together
3 * 3 * 3 = 27
5 * 5 * 5 = 125
Thus, your answer will be 27/125
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Distribute +1
78=1/2x+x-6
combine like terms
1=(2/2)
(1/2)+(2/2)= 3/2
78=3/2x-6
Add 6 to both sides
84=3/2x
Divide by (3/2)
x=$56=adult ticket
Child ticket = x-6=56-6=$50
$50
Answer:
1) (x + 3)(3x + 2)
2) x= +/-root6 - 1 by 5
Step-by-step explanation:
3x^2 + 11x + 6 = 0 (mid-term break)
using mid-term break
3x^2 + 9x + 2x + 6 = 0
factor out 3x from first pair and +2 from the second pair
3x(x + 3) + 2(x + 3)
factor out x+3
(x + 3)(3x + 2)
5x^2 + 2x = 1 (completing squares)
rearrange the equation
5x^2 + 2x - 1 = 0
divide both sides by 5 to cancel out the 5 of first term
5x^2/5 + 2x/5 - 1/5 = 0/5
x^2 + 2x/5 - 1/5 = 0
rearranging the equation to gain a+b=c form
x^2 + 2x/5 = 1/5
adding (1/5)^2 on both sides
x^2 + 2x/5 + (1/5)^2 = 1/5 + (1/5)^2
(x + 1/5)^2 = 1/5 + 1/25
(x + 1/5)^2 = 5 + 1 by 25
(x + 1/5)^2 = 6/25
taking square root on both sides
root(x + 1/5)^2 = +/- root(6/25)
x + 1/5 = +/- root6 /5
shifting 1/5 on the other side
x = +/- root6 /5 - 1/5
x = +/- root6 - 1 by 5
x = + root6 - 1 by 5 or x= - root6 - 1 by 5
