How do I solve these problems
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How do I solve these problems

[From: ] [author: ] [Date: 11-05-03] [Hit: ]
............
4n ² + 28n + 40 - 280 = 280 - 280
4n ² + 28n - 240 = 0
....................divide both sides of the equation by 4;
( 4n ² + 28n - 240 ) / 4 = 0 / 4
n ² + 7n - 60 = 0
....................factor the expression on the left;
( n + 12 ) ( n - 5 ) = 0
................................= (n)(n) + (n)(-5) + (12)(n) + (12)(-5)
................................= n ² - 5n + 12n - 60
................................= n ² + 7n - 60
....................now, either of the terms in the parenthesis could be equated to zero;
n + 12 = 0...............&...............n - 5 = 0
....................subtract 12 from both sides of the left equation, &
....................add 5 to both sides of the right equation;
n + 12 - 12 = 0 - 12...............&...............n - 5 + 5 = 0 + 5
n = - 12................&................n = 5
...................note: n cannot be negative because it would result to a negative measurement.
............................so let's just drop the left equation.

n = 5
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# 23 )))
since it is a rectangular pool, let's presume that the sides are perfectly vertical and therefore the dimensions at the bottom would be the same as that on the top of the pool.

Length = 3 + ( 2 • Width ) = 3 + 2W

area of pool = length • width
90 ft ² = ( 3 + 2W ) • W
90 = 3W + 2W ²
...................subtract 90 from both sides of the equation;
90 - 90 = 3W + 2W ² - 90
0 = 2W ² + 3W - 90
.....or.....
2W ² + 3W - 90 = 0
....................factor the expression on the left;
( 2W + 15 ) ( W - 6 ) = 0
................................= (2W)(W) + (2W)(-6) + (15)(W) + (15)(-6)
................................= 2W ² - 12W + 15W - 90
................................= 2W ² + 3W - 90
....................now, either of the terms in the parenthesis could be equated to zero;
2W + 15 = 0...............&...............W - 6 = 0
....................subtract 15 from both sides of the left equation, &
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