let a,b,c be three positive real numbers such that abc is not equal to 1, (ab)^2=(bc)^4=(ca)^x=abc, then x=?
Answers
Answer:
Therefore the relationship between x and y can be x + y = 0.
Given : If the point (x, y) is at equal distance from the x-axis and y-axis.
To find : The relationship between x and y can be .
x = 2y
x = 3y
2x + y = 0
x + y = 0
solution : a/c to question, point (x, y) is at equal distance from the x-axis and y-axis.
point on x - axis = (x, 0)
point on y - axis = (0, y)
using distance formula,
d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}d=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
distance between (x,y) and (x, 0) = distance between (x,y) and (0, y)
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2}
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x)
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0)
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0) 2
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0) 2
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0) 2
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0) 2 =
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0) 2 = (x−0)
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0) 2 = (x−0) 2
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0) 2 = (x−0) 2 +(y−y)
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0) 2 = (x−0) 2 +(y−y) 2
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0) 2 = (x−0) 2 +(y−y) 2
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0) 2 = (x−0) 2 +(y−y) 2
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0) 2 = (x−0) 2 +(y−y) 2
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0) 2 = (x−0) 2 +(y−y) 2 ⇒|y| = |x|
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0) 2 = (x−0) 2 +(y−y) 2 ⇒|y| = |x|⇒y = ± x
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0) 2 = (x−0) 2 +(y−y) 2 ⇒|y| = |x|⇒y = ± x⇒x + y = 0 or x - y = 0
istance between (x,y) and (x, 0) = distance between (x,y) and (0, y)⇒\sqrt{(x-x)^2+(y-0)^2}=\sqrt{(x-0)^2+(y-y)^2} (x−x) 2 +(y−0) 2 = (x−0) 2 +(y−y) 2 ⇒|y| = |x|⇒y = ± x⇒x + y = 0 or x - y = 0Therefore the relationship between x and y can be x + y = 0.
Step-by-step explanation: