Answer:
1/63
Step-by-step explanation:
There are a couple of ways to do this.
<h3>1) </h3>
Look for the GCF of the numerators when a common denominator is used.
GCF(3/7, 4/9) = GCF(27/63, 28/63) = (1/63)·GCF(27, 28)
GCF(3/7, 4/9) = 1/63
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<h3>2) </h3>
Use Euclid's algorithm. If the remainder from division of the larger by the smaller is zero, then the smaller is the GCF; otherwise, the remainder replaces the larger, and the algorithm repeats.
(4/9)/(3/7) = 1 remainder 1/63*
(3/7)/(1/63) = 27 remainder 0
The GCF is 1/63.
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* The quotient is 28/27 = 1 +1/27 = 1 +(1/27)(3/7)/(3/7) = 1 +(1/63)/(3/7) or 1 with a remainder of 1/63.
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<em>Additional comment</em>
3/7 = (1/63) × 27
4/9 = (1/63) × 28
Step-by-step explanation:
Answer:
A) The annual multiplier was 1.0339; the annual increase was 0.0339 of the value.
B) 3.39% per year
C) $182,000
Step-by-step explanation:
A) Let's let t represent years since 1987. Then we can fill in the numbers and solve for r.
165000 = 100000(1 +r)^15
1.65^(1/15) = 1 +r . . . . . divide by 100,000; take the 15th root
1.03394855265 -1 = r ≈ 0.0339
The value was multiplied by about 1.0339 each year.
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B) The value increased by about 3.39% per year.
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C) S = $100,000(1.03394855265)^18 ≈ $182,000
Answer:
y = 9.8x + 6.7
y = 6.7x + 9.8
y = 7.5x + 3.8
y = 3.8x + 7.5
Step-by-step explanation:
I made 2represent x.
y= 9.8(2) + 6.7
y = 26.3
y=6.7(2) + 9.8
y= 23.2
y= 7.5(2) + 3.8
y= 18.8
y= 3.8(2) + 7.5
y= 15.1
Answer:
1 hour
Step-by-step explanation:
I find these easiest to work by considering the initial difference in distance and the speed at with that gap is closing.
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The gap is 15 miles, the distance the first ship is from harbor when the second ship starts.
The rate of closure is the difference in the speeds of the two ships:
60 mph -45 mph = 15 mph
Then the closure time is ...
time = distance/speed
time = (15 mi)/(15 mi/h) = 1 h
It will take the second ship 1 hour to catch up to the first ship.
