Convert 1 3/20 to a decimal number
2 answers:
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Answer:
3 hours 28 minutes
Step-by-step explanation:
Let A = the RATE at which hose A can fill the pool alone
Let B = the RATE at which hose B can fill the pool alone
Let C = the RATE at which hose C can fill the pool alone
Hoses A and B working simultaneously can pump the pool full of water in 4 hours
From Question. The combined RATE of hoses A and B is 1/4 of the pool PER HOUR
In other words, A + B = ¼
Hoses B and C working simultaneously can pump the pool full of water in 6 hours
From Question. The combined RATE of hoses A and C is 1/4 of the pool PER HOUR
In other words, B + C = 1/6
Hoses A and C working simultaneously can pump the pool full of water in 5 hours
From Question. The combined RATE of hoses A and C is 1/5 of the pool PER HOUR
In other words, A + C = 1/5
At this point, we have the following system:
A + B = 1/4 ...1
B + C = 1/6 ...2
A + C = 1/5 ...3
Considering equation one
....4
sub equation 4 into 2
Considering equation 3
B + [B-(1/20)] = 1/6
B =7/80
Sub B Into enq 1 and 2
A =13/80
C = 3/80
A+B+C = 23/80
3.47
=> 3 hours
0.47 × 60 = 28minutes
the sequence starts at 10 and increases by 7 each time, thefore a possible sequence identification is:
Therefore, the 100th term is:
Answer:
15 years and 8 months
Step-by-step explanation:
I used the formula for compound interest as shown below and solved for the unknown, time.
Since our interest is compounded annually (So once a year) our m value is 1.
Answer:
3<em>x </em>+ 2<em>y</em> = 34. and two possible pairs of positive numbers are (<em>x</em>, <em>y</em>) = (10, 2) and (<em>x</em>, <em>y</em>) = (4, 11).
Step-by-step explanation:
Let the First positive number be <em>x</em> and second positive number be <em>y.</em>
Triple of first number = 3<em>x</em>
double of second number = 2<em>y</em>
According to question,
3<em>x </em>+ 2<em>y</em> = 34
Therefore, the equation is 3<em>x </em>+ 2<em>y</em> = 34
So the two possible pair of numbers Diego would be thinking of must satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34
Now, 3<em>x </em>+ 2<em>y</em> = 34
3<em>x </em>= 34 - 2<em>y</em>
Let x = 10 and by substituting its value in above expression,
Therefore first pair (<em>x</em>, <em>y</em>) = (10, 2)
In the same way put x = 4 then,
3<em>x </em>= 34 - 2<em>y</em>
Therefore first pair (<em>x</em>, <em>y</em>) = (4, 11)
Therefore, (<em>x</em>, <em>y</em>) = (4, 11) and (<em>x</em>, <em>y</em>) = (10, 2) are the two possible pairs of numbers Diego could be thinking of as these both values satisfy the equation 3<em>x </em>+ 2<em>y</em> = 34.