Give the equation for a circle with the given center and radius. center at (-1, 3), radius = 4

Mathematics

2 answers:

tiny-mole [99]2 years ago

7 0

Answer:

(x+1)^2 + (y-3)^2 = 16

Step-by-step explanation:

The standard form of an equation of a circle with center (h,k) and radius r is (x-h)^2 + (y-k)^2 = r^2. Plugging in those values, we get (x-(-1))^2 + (y-3)^2= 4^2 = (x+1)^2 + (y-3)^2 = 16.

Colt1911 [192]2 years ago

3 0

Answer:

(x + 1)² + (y - 3)² = 16

Step-by-step explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r is the radius

here (h, k ) = (- 1, 3 ) and r = 4 , then

(x - (- 1) )² + (y - 3)² = 4² , that is

(x + 1)² + (y - 3)² = 16

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Which function has a range of y < 3?<br> y=3(2)*<br> y=2(3)<br> y=-(2)*+ 3<br> O y=(2]*_3

jasenka [17]

Answer:

The function given by $y = - (2)^{x} + 3$ will have a range of y < 3.

Step-by-step explanation:

The function given by $y = - (2)^{x} + 3$ will have a range of y < 3.

This is because, for any real values of x, the term $- (2)^{x}$ will have a negative value. If we put x = 1, then, $- (2)^{x}$ = - 2, and for x = -1, then $- (2)^{x}$ = - 0.5.