12 - 3b < 9

12 (-12) - 3b < 9 (-12)

-3b < -3

-3b/-3 < -3/-3

(note when dividing a negative number, flip the inequality sign)

b > 1 is your answer

hope this helps

3 0

3 years ago

Write the equation of a circle whose center is at (3,-2) and has a radius of 11.<br>Show your work!

Viefleur [7K]

Answer:

$(x - 3)^{2} + (y + 2)^{2} = 121$

Step-by-step explanation:

The equation of a circle has the following format:

$(x - x_{0})^{2} + (y - y_{0})^{2} = r^{2}$

In which r is the radius and the centre is the point $(x_{0}, y_{0})$

In this question:

Center at (3,-2), so $x_{0} = 3, y_{0} = -2$

Radius 11, so $r = 11$

Then

$(x - 3)^{2} + (y - (-2))^{2} = 11^{2}$

$(x - 3)^{2} + (y + 2)^{2} = 121$

3 0

3 years ago

Help me asap please let me down

stiv31 [10]

I would divide 120 by 6 and get 20

3 0

3 years ago

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