C) the left side of the equation should be -15 + 3r after distributing.
Pay attention here because I'm adding an extra letter to our circle to help keep track of the values in our formula. OUTSIDE of the intercepted arc I'm adding the point E. So the major arc is arc BEG and the minor arc is arc BG. The formula then for us is ∠

. We just don't have values for the arcs yet. If the measure of the central angle is 4x+238, then the measure of arcBG is also 4x+238. Around the outside of the circle is 360°. So we will use it in an expression. ArcBEG=360-(4x+238). Fitting that into our formula we have
![2x+146= \frac{1}{2}[(360-4x-238)-(4x+238)]](https://tex.z-dn.net/?f=2x%2B146%3D%20%5Cfrac%7B1%7D%7B2%7D%5B%28360-4x-238%29-%284x%2B238%29%5D%20)
. Doing all the simplifying inside there we have

and

. Multiply both sides by 2 to get rid of the fraction: 4x+292=-8x-116. Combine like terms to get 12x = -408 and divide to solve for x. x = -34. Fourth choice down from the top.
Answer:
(mn+n²)/(m+n)
Step-by-step explanation:
probability of blue marble= n/(n+m)
probability of red marble= m/(n+m)
probability that process stops = Probability of both blue + probability of both red= n/(n+m) × n/(n+m) + m/(n+m)×m/(n+m)
= (n²+m²)/(n+m)²
P(1st marbel is blue)= P(blue and red) + P(blue and blue)
= mn/(n+m) + n²/(n+m)
= (mn+n²)/(m+n)
P(1st marble is blue | process stops)= ( (mn+n²)/(m+n)× (n²+m²)/(n+m)²)/ ((n²+m²)/(n+m)²)
= (mn+n²)/(m+n)
Answer:
2.9 mi
Step-by-step explanation:
The time difference t between 12.30pm and 3.30pm is 3h.
Let the distance to Marat's house be s.
The equation to calculate velocity v is given by:
v = distance / time.
Now you can write an equation for the time difference t and use the two velocities and the 90 min(= 1.5h) she stays:
3 = s/12 + 1.5 + s/2.3
Solve this equation for s:
(1/12 + 10/23)* s = 1.5
s = 2.9
12 + 26 + 125 = 163
163 + 0.056(163) = 163 + 9.13 = 172.13