A box contains ten cards labeled Q, R, S, T, U, V, W, X, Y, and Z. One card will be randomly chosen.
1 answer:
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Answer: (12) ∠1 = 20° (13) ∠2 = 50° (14) ∠3 = 15° (15) UV = 80° (16) AB = 40° (17) ABC <em>or</em> 180° - CD (18) BC - 140° (19) ABC = 150°
<u>Step-by-step explanation:</u>
12) (UV - ST) = ∠1
(80 - 40) = ∠1
(40) = ∠1
20 = ∠1
13) (UV + ST) = ∠2
(70 + 30) = ∠2
(100) = ∠2
50 = ∠2
14) (VB - BS) = ∠3
(60 - 30) = ∠3
(30) = ∠3
15 = ∠3
15) (UV - ST) = ∠1
(UV - 20) = 30
UV - 20 = 60
UV = 80
16) ∠1 = arc AB
∠1 = 40
arc AB = 40
17) arc AB + arc BC = arc AC
= 180 = arc CD
18) ∠1 + ∠2 + ∠3 = 180
20 + ∠2 + 20 = 180
∠2 + 40 = 180
∠2 = 140
19) ∠1 + ∠ 2 = arc ABC
∠1 + ∠2 + ∠3 = 180
arc ABC + 30 = 180
arc ABC = 150
Answer:
It would take the newer pump 4.5 hours to drain the pool
Step-by-step explanation:
Let's investigate first what is the fraction of the job done in the unit of time (hour in this case) by each pump if the work individually:
older pump: if it takes it 9 hours to complete the job, it does of the job in one hour.
newer pump: we don't know how long it takes (this is our unknown) so we call it "x hours". Therefore, in the unit of time (in one hour) it would have completed of the total job.
both pumps together: since it takes both 3 hours to complete the job, in one hour they do of the job.
Now, we can write the following equation about fractions of the job done:
<em>The fraction of the job done by the older pump plus the fraction of the job done by the newer pump in one hour should total the fraction of the job done when they work together.</em> That is in mathematical terms:
and solving for x: