please help me to solve that questions
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.
(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.
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.
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.
(x -2)^2 + (y+3)^2 = 144 is the desired circle.