suppose the equations (x-4)^2 + (y-3)^2=4 and (x-4)^2/4 + (y-10)^2/16=1 are graphed in the same standard (x, y) coordinate plan how many points of intersection

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb

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.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: