there is two questions that I need help with:
The variable x represents a real number between 10 and 20 inclusive.The variable Y represents a real number between .2 and 5,inclusive.What is the greatest possible integer value of the quotient x/y?
2nd question: A square with an area of 36 units is drawn in quadrant II of a coordinate grid.One of the vertices is at the origin.What is the x-coordinate of the vertex farthest from the origin?
The largest quotient can be obtained when the numerator is largest and the denominator is smallest
so since in x/y, x goes from 10 to 20
and y goes from .2 to 5
the largest quotient is 20/.2 which is 100
the farthest vertex would be (-x,y) , the others would be (-x,0) , (0,y) and (0,0)
and xy = 36
where x = 36/y
We will have to restrict our x and y to integers, or else we have no answer.
e.g. if y = .01, then x = 36/.01 = 3600
and the point is (-3600 , .01)
and the point would be √(3600^2 + .01^2)
or 3600 units away
if I make y smaller, then x will become even larger, and the point would get farther and farther away.
so....
the smallest value I can choose for y is 1
then x = 36/1 = 36
and our farthest point is (-36,1)
(it also could have been (1,36) of course )
#2. I believe we are discussing a square, not any rectangle.
x = -6
y = 6
Let's start with the first question:
To find the greatest possible integer value of the quotient x/y, we need to determine the maximum value of x and the minimum value of y within the given ranges.
The variable x represents a real number between 10 and 20 inclusive. So the maximum value of x is 20.
Similarly, the variable y represents a real number between 0.2 and 5 inclusive. So the minimum value of y is 0.2.
Now, to find the greatest possible integer value of x/y, we divide the maximum value of x by the minimum value of y:
20 / 0.2 = 100
Therefore, the greatest possible integer value of the quotient x/y is 100.
Moving on to the second question:
We know that a square with an area of 36 units is drawn in quadrant II of a coordinate grid, and one of the vertices is at the origin (0, 0). We need to find the x-coordinate of the vertex farthest from the origin.
Since the square is in quadrant II, the x-coordinate will be negative. Let's assume the coordinates of the vertices are (x, y).
The area of a square is given by the formula: A = side^2
In our case, the area is 36 units. So, the side of the square is √36 = 6 units.
Since the x-coordinate is negative, the farthest vertex from the origin will have the largest absolute value of the x-coordinate. We need to find the vertex with a y-coordinate of 6 units.
Using the Pythagorean theorem, we know that the distance from the origin to any vertex of the square is equal to the length of its side (6 units).
Let's assume the vertex with a y-coordinate of 6 units has an x-coordinate of -a (negative).
Using the Pythagorean theorem:
(-a)^2 + 6^2 = 6^2
a^2 + 36 = 36
a^2 = 0
a = 0
So, the x-coordinate of the farthest vertex from the origin is -0 = 0.
Therefore, the x-coordinate of the vertex farthest from the origin is 0.