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my answer is 15

2006-08-13 09:44:25 · 12 answers · asked by Retta 1 in Education & Reference Homework Help

12 answers

Looks like 72 to me...

2006-08-13 11:16:29 · answer #1 · answered by 42ITUS™ 7 · 0 0

W=2

2006-08-13 09:51:59 · answer #2 · answered by Brian 3 · 0 0

2

2006-08-13 09:49:21 · answer #3 · answered by Anonymous · 0 0

w=2

2006-08-13 09:51:13 · answer #4 · answered by Anonymous · 0 0

Change by substitution

square root of 3 = 15 - 9 (6)
Thus 3W = 6 x 6 (36)
W = 36/3 (12)

2006-08-13 09:50:25 · answer #5 · answered by quatt47 7 · 0 0

3w+9=15

1. First, you subtract 9 from both sides.

3w=6

2. Divide each side by 3.

w=2

3. Find the square root of both sides.

The square root of w=Positive square root of 2 or Negative square root of 2

2006-08-13 09:54:18 · answer #6 · answered by Darkwings 2 · 0 0

72 is correct. First square both sides of the eqn to get rid of the radical. This gives you 3w+9=225. Subtract 9 from both sides.
This results in 3w=216. Then divide both sides by 3, isolating the variable. w=72.

2006-08-13 13:54:41 · answer #7 · answered by curious 3 · 0 0

Is the square root over 3w, or the whole quantity (3w+9).

If it's just over 3w:

sqrt(3w) =6 by subtracting 9 from both sides

Square both sides: 3w = 36

Divide by 3: w=12 **********


BUT, if the sqrt covers the whole quantity 3w+9, start by squaring both sides first:

3w+9 = 225 Then sub. 9 from both sides:

3w = 216

Div. by 3: w=72 ****************

2006-08-13 10:36:13 · answer #8 · answered by jenh42002 7 · 0 0

the square root of 2 (approx 1.414)

3w+9=15

3w=15-9

3w=6

w=6/3

w=2

sq.root of 2 = approx 1.414243562

2006-08-13 09:52:09 · answer #9 · answered by EricC 2 · 0 0

First step minus 9 from both sides of the equal sign which gives you 3w=6.

Secondly isolate the w by dividing 3 by both sides getting the answer w=5.

I hope that my answer helps. :)

2006-08-13 12:04:12 · answer #10 · answered by zain 2 · 0 0

w=72

(is the square root bar over all of '3w+9'?)

2006-08-13 09:51:10 · answer #11 · answered by Anonymous · 0 0

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