Answer:
To find a
sin 30=a/6in
1/2=a/6in
2a=6in
a=6in/2
a=3in
To find b
cos 30=b/6in
/3/2=b/6in
2b=/3*6in
b=2/3(two radical three)
I don't get the radical sign so I use / sign on the last 3 steps on the second
First solve for the other angles.
You know one angle is 15º.
The one across from it is also 15º since it is a vertical angle to the known 15º angle.
Both of those together gets 30º.
The whole thing is 360º, so just subtract 30 from 360 and divide.
360 - 30 = 330
Do the same for the 15x angle.
The one under/across to it is a vertical angle so is also 15xº.
Now you are left with 30xº = 330.
Just divide now.
330 / 30 = 11
x = 11
15x = 165
Checking your work:
15x = 165
15x = 165
15
15
Those are the angles.
Add those numbers together and make sure it adds up to 360º.
165 + 165 + 15 + 15 = 360
So this is correct.
Subtract 510 from 375 because you need to find how many more points that he needs to get
Suppose the larger pump alone can empty the tank in L hours, and the smaller pump can finish the job in S hours, then each hour the large pump empties 1/L portion of the tank, and the small pump empties 1/S per hour
Working together for three hours, they empty the whole tank, which is 100% of it, so 3/L+3/S=100%=1
Larger pump can empty the tank in 4 hours less than the smaller one, so L=S-4
replace L: 3/(S-4)+3/S=1
Make the denominator the same to solve for:
3S/[S(S-4)] +3(S-4)/[S(S-4)]=1
(3S+3S-12)/[S(S-4)]=1
(3S+3S-12)=[S(S-4)]
S^2-10s+12=0
use the quadratic formula to solve for S
S is about 8.6
The answer is not whole hour.
