Can I have some help with this? Construct a linear function that has a slope of -3 and passes through the point (-2,3).

-3x-3

Step-by-step explanation:

We want our equation to look like this

y=mx+b

we know that m= -3 so we just need to find b

We can start by plugging in the ordered coordinates into our equation

3=(-3)(-2)+b

Solve for b

3=6+b

b= -3

Which means our final equation is

y= -3x-3

x = central angle apc = measure of minor arc ac

the two angles x and 30 add to 180, so x+30 = 180 and x = 180-30 = 150. the measure of minor arc ac is 150 degrees.

what we need to do from here is find the circumference and then multiply it by x/360 since there are 360 degrees in a circle. in this case, x = 150, so x/360 = 150/360 = 15/36 = 5/12. the shortest distance along the curve from a to c is 5/12 of the distance around the entire circle (aka circumference)

c = circumference

c = pi*d

c = pi*24

c = 24pi

now multiply this by 5/12 to get

distance along minor arc ac = (5/12)*(circumference)

distance along minor arc ac = (5/12)*(24pi)

distance along minor arc ac = 10pi

the formula you can use is

l = (x/360)*pi*d

with

l = length of minor arc

x = central angle

d = diameter

which essentially is the steps we took though a bit differently.

10000

step-by-step explanation: divide by 2,