please help me to solve that questions

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- Oct 12th 2006, 10:16 AM #1

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- Jul 2006
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- Oct 12th 2006, 01:26 PM #2
1. |(2/x) - 4| < 3

This is really two inequalities:

-3 < (2/x) - 4 < 3

So add 4 to all "three" sides:

-3 + 4 < (2/x) - 4 + 4 < 3 + 4

1 < 2/x < 7

So 1 < 2/x and 2/x < 7

The first gives:

0 < 2/x -1

0 < (2-x)/x

which is positive for x < 2.

The second gives:

2/x < 7

2/x - 7 < 0

(2-7x)/x < 0

which is negative for x > 2/7.

So 2/7 < x < 2.

-Dan

- Oct 12th 2006, 01:30 PM #3
2. Given points: (3, 4), (5, 8) and (3, 9)

Show that they form the points of a right triangle.

I'm not going to do all of it, just point you along.

Slopes: If this is a right triangle then two lines connecting these points are perpendicular. Two lines are perpendicular if m1*m2 = -1, where m1 and m2 are the slopes of the lines in question. So pairing off the points, find the slopes of the lines between them (you'll have 3 lines). Now determine if any two of these are perpendicular.

Distances: If this is a right triangle, then the distances will follow a^2 + b^2 = c^2, where c is the longest side. So find the distance between each pair of points and see if it fits the Pythagorean Theorem.

-Dan

- Oct 12th 2006, 01:33 PM #4
(1)Find the equation of line which has y intercept “-5” and is perpendicular to the line joining the points (3, 6) and (1, 8)

The equation of your line is

y = mx + b.

We know that the y-intercept is -5, so b = -5

Now, we know this line is perpendicular to the line connecting the points (3,6) and (1,8). Call this slope M. We know then that m*M = -1. So find M, then you can find m.

-Dan

- Oct 12th 2006, 01:36 PM #5

Find the equation of the circle with centre at (2, -3) and whose radius is “3” times the radius of x^2 + y^2 - 2x - 4y = 11.

The first thing we need to do is put the second circle into standard form. We do this by completing the square for both x and y:

x^2 - 2x + 1 + y^2 - 4y + 4 = 11 + 1 + 4

(x - 1)^2 + (y - 2)^2 = 16

So this circle has a radius of 4. The circle we want has a radius of 3*4 = 12.

So

(x -2)^2 + (y+3)^2 = 144 is the desired circle.

-Dan