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3 years ago

12

Center (3, -2) Radius = 3

1 answer:

mel-nik [20]3 years ago

3 0

Answer:

x^2-6x+y^2+4y=-4

Step-by-step explanation:

I am assuming you want the equation of the circle. We know that the equation of a circle is (x-h)^2 + (y-k)^2 =r^2, where r is the radius, and h and k are the x and y coordinates of the center respectively. We can solve for this by plugging in the numbers.

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

x^2-6x+9+y^2+4y+4=9

x^2-6x+y^2+4y=-4

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Find the slope of the line which is perpendicular to a line having a slope of 5/7

erica [24]

Answer:

-7/5

Step-by-step explanation:

5/7

= -7/5

3 0

2 years ago

Which series of transformations will not map figure H onto itself

7nadin3 [17]

Answer:

D

Step-by-step explanation:

Given a square with vertices at points (2,1), (1,2), (2,3) and (3,2).

Consider option A.

1st transformation $(x+0,y-2)$ will map vertices of the square into points

$(2,1)\rightarrow (2,-1);$
$(1,2)\rightarrow (1,0);$
$(2,3)\rightarrow (2,1);$
$(3,2)\rightarrow (3,0).$

2nd transformation = reflection over y = 1 has the rule (x,2-y). So,

$(2,-1)\rightarrow (2,3);$
$(1,0)\rightarrow (1,2);$
$(2,1)\rightarrow (2,1);$
$(3,0)\rightarrow (3,2)$

These points are exactly the vertices of the initial square.

Consider option B.

1st transformation $(x+2,y-0)$ will map vertices of the square into points

$(2,1)\rightarrow (4,1);$
$(1,2)\rightarrow (3,2);$
$(2,3)\rightarrow (4,3);$
$(3,2)\rightarrow (5,2).$

2nd transformation = reflection over x = 3 has the rule (6-x,y). So,

$(4,1)\rightarrow (2,1);$
$(3,2)\rightarrow (3,2);$
$(4,3)\rightarrow (2,3);$
$(5,2)\rightarrow (1,2)$

These points are exactly the vertices of the initial square.

Consider option C.

1st transformation $(x+3,y+3)$ will map vertices of the square into points

$(2,1)\rightarrow (5,4);$
$(1,2)\rightarrow (4,5);$
$(2,3)\rightarrow (5,6);$
$(3,2)\rightarrow (6,5).$

2nd transformation = reflection over y = -x + 7 will map vertices into points

$(5,4)\rightarrow (3,2);$
$(4,5)\rightarrow (2,3);$
$(5,6)\rightarrow (1,2);$
$(6,5)\rightarrow (2,1)$

These points are exactly the vertices of the initial square.

Consider option D.

1st transformation $(x-3,y-3)$ will map vertices of the square into points

$(2,1)\rightarrow (-1,-2);$
$(1,2)\rightarrow (-2,-1);$
$(2,3)\rightarrow (-1,0);$
$(3,2)\rightarrow (0,-1).$

2nd transformation = reflection over y = -x + 2 will map vertices into points

$(-1,-2)\rightarrow (4,3);$
$(-2,-1)\rightarrow (3,4);$
$(-1,0)\rightarrow (2,3);$
$(0,-1)\rightarrow (3,2)$

These points are not the vertices of the initial square.

5 0

3 years ago

Multiply two and five-eighths negative two and three-fifths .

Yuki888 [10]

The answer is -6.825 in fraction form, it would be -6 33/40

4 0

3 years ago

Make vthe subjject of the formulaE=mvsquare/ 2<br>find the value of v when m =2 and E=64

Maksim231197 [3]

E=mv²

divide both sides by m

E/m=v²

square root both sides

v=√E/m

m=2,E=64

v=√64/2

v=9/√2

6 0

3 years ago

Find three cases consecutive integers whose sum is 363

Scilla [17]

Answer:

120,121,122

Step-by-step explanation:

4 0

3 years ago