1.determine whether this is a linear equation and give the solution:
a.y= -5
2.determine if a perpendicular line or parallel line and please give the solution:
a. x= -y + 9 and y= -x-2
3.find the distance between the given points and include the solution:
a. (-3,7),(5, -4)
4.find the intersection of a line y=5-2x and the circle x^2 +y^2 = 5.
5.describe the graph:
a.the point (x,y) is less than or equal to 5 units from the center (3,2) of a circle.
thank you very much for the answers...these are the only ones i can't answer..thank you very much!:):)
a.y= -5
2.determine if a perpendicular line or parallel line and please give the solution:
a. x= -y + 9 and y= -x-2
3.find the distance between the given points and include the solution:
a. (-3,7),(5, -4)
4.find the intersection of a line y=5-2x and the circle x^2 +y^2 = 5.
5.describe the graph:
a.the point (x,y) is less than or equal to 5 units from the center (3,2) of a circle.
thank you very much for the answers...these are the only ones i can't answer..thank you very much!:):)
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1.determine whether this is a linear equation and give the solution:
a.y= -5
no matter what the value of x, y will always = -5. so this is a linear equation (or straight line) parallel to the x-axis at y = -5.
2.determine if a perpendicular line or parallel line and please give the solution:
a. x= -y + 9 and y= -x-2
parallel lines have the same gradient.
x= -y + 9 .....(i)
y = -x +9
y= -x- 2.......(ii)
both (i) and (ii) have the same gradient (-1) ...the coeffiecent (number) in front of the x.
They are parallel
3.find the distance between the given points and include the solution:
a. (-3,7),(5, -4)
d = √ ((diff in y)² + (diff in x)²)
d = √ ((11)² + (8)²)
d = √ (121 + 64)
d = 13.6
4.find the intersection of a line y=5-2x and the circle x^2 +y^2 = 5.
substitute y=5-2x in x^2 +y^2 = 5
x² +(5- 2x)² = 5
x² + 25 -20x +4x² = 5
x² + 20 -20x +4x² = 0
x² + 25 -20x +4x² = 5........simplifying
5x² -20x + 25 = 0
x² -4x + 5 = 0
no factors - line does not cut circle
5.describe the graph:
a.the point (x,y) is less than or equal to 5 units from the center (3,2) of a circle.
thank you very much for the answers...these are the only ones i can't answer..thank you very much!:):)
Bit hard to envisage without a radius to the circle. Put the point (3,2) on a graph and use a compass to draw a circle of radius 5 units around that point then shade the circle indicating (x,y) could be anywhere in that shaded area (including the line drawn).
a.y= -5
no matter what the value of x, y will always = -5. so this is a linear equation (or straight line) parallel to the x-axis at y = -5.
2.determine if a perpendicular line or parallel line and please give the solution:
a. x= -y + 9 and y= -x-2
parallel lines have the same gradient.
x= -y + 9 .....(i)
y = -x +9
y= -x- 2.......(ii)
both (i) and (ii) have the same gradient (-1) ...the coeffiecent (number) in front of the x.
They are parallel
3.find the distance between the given points and include the solution:
a. (-3,7),(5, -4)
d = √ ((diff in y)² + (diff in x)²)
d = √ ((11)² + (8)²)
d = √ (121 + 64)
d = 13.6
4.find the intersection of a line y=5-2x and the circle x^2 +y^2 = 5.
substitute y=5-2x in x^2 +y^2 = 5
x² +(5- 2x)² = 5
x² + 25 -20x +4x² = 5
x² + 20 -20x +4x² = 0
x² + 25 -20x +4x² = 5........simplifying
5x² -20x + 25 = 0
x² -4x + 5 = 0
no factors - line does not cut circle
5.describe the graph:
a.the point (x,y) is less than or equal to 5 units from the center (3,2) of a circle.
thank you very much for the answers...these are the only ones i can't answer..thank you very much!:):)
Bit hard to envisage without a radius to the circle. Put the point (3,2) on a graph and use a compass to draw a circle of radius 5 units around that point then shade the circle indicating (x,y) could be anywhere in that shaded area (including the line drawn).
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1. Yes, the line is a horizontal line going through -5 on the Y axis.
2. Parallel, since if you solve both for y, x is the same. y= -x + 9 and y = -x - 2. The lines are 45 degree angle diagonal lines, one going through -2 on the y axis and one going through 9 on the y axis.
3. Distance formula:
d = (y2 - y1)/(x2 - x1)
d = (7 - (-4))/(5 - (-3)) = 11/8
d = 11/8
4. substitute (5-2x) for the y in the circle equation and solve for x, then plug x into the other equation.
5. I'm not sure what this question is.
2. Parallel, since if you solve both for y, x is the same. y= -x + 9 and y = -x - 2. The lines are 45 degree angle diagonal lines, one going through -2 on the y axis and one going through 9 on the y axis.
3. Distance formula:
d = (y2 - y1)/(x2 - x1)
d = (7 - (-4))/(5 - (-3)) = 11/8
d = 11/8
4. substitute (5-2x) for the y in the circle equation and solve for x, then plug x into the other equation.
5. I'm not sure what this question is.