If the example is y= -2x and y= x+3 Does the point (2,5) make either equation true?
2 answers:
You might be interested in
Answer: The answer is 6.
The number sentence is
3-3=0
10-4=6
So, 13-7=6
Step-by-step explanation:
To solve , by making ten
First break 13 and write
Then break 7 as
Now to subtract 7 from 13,
Step 1. Subtract 3 (a part of 7), from 3(a part of 13) or cross 3 circles, then we have left 10-4
Step 2. Subtract 4 from 10 or cross 4 circles out of 10, we get 6.
The answer is 6.
The standard equation of a circle is
(x-h)^2 + (y-k)^2 = r^2
where the center is at point (h,k)
From the statement of the problem, it is already established that h = 2 and k = -5. What we have to determine is the value of r. This could be calculated by calculating the distance between the center and point (-2,10). The formula would be
r = square root [(x1-x2)^2 + (y1-y2)^2)]
r = square root [(2--2)^2 + (-5-10)^2)]
r = square root (241)
r^2 = 241
Thus, the equation of the circle is
(x-h)^2 + (y-k)^2 = r^2
where the center is at point (h,k)
From the statement of the problem, it is already established that h = 2 and k = -5. What we have to determine is the value of r. This could be calculated by calculating the distance between the center and point (-2,10). The formula would be
r = square root [(x1-x2)^2 + (y1-y2)^2)]
r = square root [(2--2)^2 + (-5-10)^2)]
r = square root (241)
r^2 = 241
Thus, the equation of the circle is