If the equation of a circle is (x+5)^2 + (y-7)^2=36, its center point is

1 answer:

3 0

Answer:

centre = (- 5, 7 )

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

(x + 5)² + (y - 7)² = 36 ← is in standard form

with h = - 5 and k = 7

centre = (- 5, 7)

You might be interested in

Vinil7 [7]

<u>Answer:</u>

$\frac{11}{31}$

<u>Step-by-step explanation:</u>

We are given the following expression and we are to perform operation and write the answer in simplest form:

$\frac { 2 \frac { 1 } { 5 } } { \frac { 31 } { 5 } }$

Changing the mixed fraction into improper fraction:

$2\frac{1}{5}$ ---> $\frac{11}{5}$

Now simplifying it:

$\frac { \frac { 1 1 } { 5 } } { \frac { 31 } { 5 } }$

Changing the division sign to multiplication by taking the reciprocal of the denominator:

$\frac{11}{5} \times \frac{5}{31}$

$\frac{11}{31}$

5 0

3 years ago

hodyreva [135]

Answer:

$q = 16 \sqrt{2}$ and $r = 16$

Step-by-step explanation:

We know that

$\sin( \alpha ) = \frac{side \: opposite \: to \: \alpha }{hypotenuse \: of \: triangle}$
$\cos( \alpha ) = \frac{side \: adjacent \: to \: \alpha }{hypotenuse \: of \: traingle}$
$\tan( \alpha ) = \frac{side \: opposite \: to \: \alpha }{side \: adjacent \: to \: \alpha }$