Math Questions!!!!!?

Question

1)AreAll real numbers a solution to the equation: 6x - 9 = -3(3 – 2x)

2)6(2x + 4) = 3(8 + 4x)

3)5x + 3(x - 2) = 4x - (x - 4)

4)-5(12 - 4x) = 8(3x + 9)

5)Absolute value equation:

|x + 3| - 4 = 12

Answer 1

1) 6x-9=-3(3-2x)
6x-9=-9+6x
since both sides are equal hence all real numbers are a solution for this
2)6(2x + 4) = 3(8 + 4x)
12x+24=24+12x
once again both sides are equal so allreal numbers are the solution for this equation
3)5x+3(x-2)=4x-(x-4)
5x+3x-6=4x-x+4
8x-6=3x+4
5x=4+6
x=2, is the only solution of this equation
4)5(12 - 4x) = 8(3x + 9)
60-20x=24x+72
-20x-24x=72-60
-44x=12
x=-3/11 is the only solution of the equation
5)|x + 3| - 4 = 12
taking the positive value of the mod first
x+3-4=12
x-1=12
x=13
taking the negative value of mod now
-(x+3)-4=12
-x-3-4=12
-x=12+3+4
x=-19
so x can take values 13 or -19

Answer 2

1) 6x - 9 = -3(3 - 2x)
<=> 6x - 9 = -3(3) - (-3)(2x)
<=> 6x - 9 = -9 + 6x
which is an identity, so true for all x.

2) 6(2x + 4) = 3(8 + 4x)
<=> 6(2x) + 6(4) = 3(8) + 3(4x)
<=> 12x + 24 = 24 + 12x
which is an identity, so true for all x.

3) 5x + 3(x - 2) = 4x - (x - 4)
<=> 5x + 3x + 3(-2) = 4x - x - (-4)
<=> 8x - 6 = 3x + 4
<=> 8x - 3x = +6 + 4
<=> 5x = 10
<=> x = 2.

4) -5(12 - 4x) = 8(3x + 9)
<=> -5(12) + (-5)(-4x) = 8(3x) + 8(9)
<=> -60 + 20x = 24x + 72
<=> 20x - 24x = +60 + 72
<=> -4x = 132
<=> x = 132/(-4) = -33

5) |x + 3| - 4 = 12
<=> |x + 3| = 16
<=> x + 3 = 16 or x + 3 = -16
<=> x = 13 or x = -19.

Answer 3

1) (6x-9)/(-3) =(3-2x)
-2x+3 = 3-2x
-4x = 0
x= 0


2) (6/3) (2x+4) = (8+4x)
2(2x+4) = 4x+8
4x+8 = 4x+8
x=0

3) 5x + 3x - 6 = 4x-x+4
8x-3x = 4+6
5x = 10
x = 2
4) -60+20x = 24x + 72
20x-24x = 72 + 60
-4x = 132
x = -33
5) |x+3| = 16
-x+3 = 16 AND x+3= 16
x= -13 OR x= 13
x= -13 does not work, therefore, the answer is x=13

Answer 4

1) 6x-9=-9+6x
0=0

2) 12x+24=24+12x
0=0

3) 5x+3x-6=4x-x+4
8x-3x =10
x=2

4) -60+20x=24x+72
20x-24x =72+60
-4x =132
x = -33

5) x+3-4=12
x =12+1
x =13

I hope it helps you.

Answer 5

I see your combination up - A = PiR^2 appropriate? nicely, area = Pi (radius squared), the place you placed the radius as 40 8. The Diameter is 24 cm, appropriate? nicely, interior the formulation, radius is barely a million/2 of the diameter, so which you may desire to be utilising 12^2, no longer 40 8^2 Your formulation might desire to look like this: area = Pi(3.14) * 12^2 That'll supply you the outstanding answer.

Answer 6

1. Yes

2. Yes

3. No

4. No

5. x = 13 and -19

Answer 7

1) yes
2) no
3) no
4) no
5) i think x should be changed to negative-sorry i don't fully understand the question in the last one!!! hope these answers are correct!!!!!

Answer 8

1. yes

2. yes

3. no

4. no

5. yes