The equation of the new circle will be,
⇒ (x - h)² + (y - k)² = r².
The circle is a closed two dimensional figure , in which the set of all points is equidistance from the center.
Now, If the center of the circle were moved from the origin to the point (h, k) and point P at (x, y) remains on the edge of the circle.
Then, Equation of the new circle,
(x - h)² + (y - k)² = r²
Where center = (h, k)
radius = r
Hence, the equation of the new circle will be,
⇒ (x - h)² + (y - k)² = r².
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Answer:(x-h)^2+(y-k)^2=r^2
Step-by-step explanation:
I just answered it and it’s correct
-4x+6y+7
Answers
A- (5/4, 2)
B- (2, 5/4)
C- (4/5,2)
D (2, 4/5)
Answer:
5/4, 2 or A on edge 2021
Step-by-step explanation:
took the test :)
6y^3 − 9y^2
Answer:
3y²(2y - 3)
Step-by-step explanation:
Given
6y³ - 9y² ← factor out 3y² from each term
= 3y²(2y - 3)
G. 3a + 3c
H. 2a + c
J. a + 2c
K. ac + a^2
The required expression that has an even integer value for all integers a and c is F. 8a + 2ac.
Simplification involves applying rules of arithmetic and algebra to remove unnecessary terms, factors, or operations from an expression.
Here,
We can determine which expressions have an even integer value for all integers a and c by looking at their factors.
An integer is even if and only if it is divisible by 2. Therefore, an expression will have an even integer value for all integers a and c if and only if it contains a factor of 2.
Let's look at each of the given expressions:
F. 8a + 2ac = 2(4a + ac)
This expression has a factor of 2, so it has an even integer value for all integers a and c.
G. 3a + 3c = 3(a + c)
This expression does not have a factor of 2, so it does not have an even integer value for all integers a and c.
H. 2a + c
This expression does not have a factor of 2, so it does not have an even integer value for all integers a and c.
J. a + 2c
This expression does not have a factor of 2, so it does not have an even integer value for all integers a and c.
K. ac + a²
This expression does not have a factor of 2, so it does not have an even integer value for all integers a and c.
Therefore, the expression that has an even integer value for all integers a and c is F. 8a + 2ac.
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Answer:
-11
Step-by-step explanation: