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How many leaps must the greyhound take to catch the hare? Task from the Elements of Algebra by L. Euler
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Find the ratio of speed?How much time will the pipe take?Calculate how long does it take to complete a task by two workersHow are we to correct this error?(from Gelfand's Algebra)Simplify expression $-ln(2)2xleft(frac12right)^x^2+1$How many minutes does Train $1$ take for the entire trip?Formula for calculating the total number of solution pairs to an equationHow many football fans were originally travelling to the game?Finding the compound interest. Task from the Elements of Algebra by L. EulerRequired in what time cistern would be filled by the second pipe alone. Task from the Elements of Algebra by L. Euler
$begingroup$
A hare is 50 leaps before a greyhound, and takes 4 leaps to the greyhound’s 3; but two of the greyhound’s leaps are as much as three of the hare’s.
My solution was $4(l + 50) = 3*frac32l$. But after calculating the answer doesn't seem to be right. I can't understand what am I missing.
algebra-precalculus
asked Mar 31 at 19:45
JohnyJohny
134
$endgroup$
add a comment |
$begingroup$
A hare is 50 leaps before a greyhound, and takes 4 leaps to the greyhound’s 3; but two of the greyhound’s leaps are as much as three of the hare’s.
My solution was $4(l + 50) = 3*frac32l$. But after calculating the answer doesn't seem to be right. I can't understand what am I missing.
algebra-precalculus
asked Mar 31 at 19:45
JohnyJohny
134
$endgroup$
add a comment |
$begingroup$
A hare is 50 leaps before a greyhound, and takes 4 leaps to the greyhound’s 3; but two of the greyhound’s leaps are as much as three of the hare’s.
My solution was $4(l + 50) = 3*frac32l$. But after calculating the answer doesn't seem to be right. I can't understand what am I missing.
algebra-precalculus
asked Mar 31 at 19:45
JohnyJohny
134
$endgroup$
A hare is 50 leaps before a greyhound, and takes 4 leaps to the greyhound’s 3; but two of the greyhound’s leaps are as much as three of the hare’s.
My solution was $4(l + 50) = 3*frac32l$. But after calculating the answer doesn't seem to be right. I can't understand what am I missing.
algebra-precalculus
algebra-precalculus
asked Mar 31 at 19:45
JohnyJohny
134
asked Mar 31 at 19:45
JohnyJohny
134
asked Mar 31 at 19:45
JohnyJohny
134
asked Mar 31 at 19:45
JohnyJohny
134
asked Mar 31 at 19:45
JohnyJohny
134
134
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The rabbit jumps $j_r$ ft in a leap.
The dog jumps $j_d$ ft in a leap.
$2j_d = 3j_r $.
The dog runs at 3j_d per unit time.
The rabbit runs at 4j_r per unit time
The dog starts at d(0) = 0.
The rabbit starts at $r(0) = d(0) + 50j_r. $
$d(t) = 0 + 3j_dt.$
$r(t) = 50j_r + 4j_rt.$
Solve $d(t) = r(t) $ for t.
How far did the dog run to catch up with the rabbit?
How many leaps did the dog make?
But alas, at the last moment, the rabbit ran into a rabbit hole.
The dog ran after the rabbit into the hole and got stuck.
The rabbit ran out the rabbit hole at the other side of the borrow.
The dog had to dig himself out. The rabbit had to make repairs.
answered Mar 31 at 23:08
William ElliotWilliam Elliot
9,1812820
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
The rabbit jumps $j_r$ ft in a leap.
The dog jumps $j_d$ ft in a leap.
$2j_d = 3j_r $.
The dog runs at 3j_d per unit time.
The rabbit runs at 4j_r per unit time
The dog starts at d(0) = 0.
The rabbit starts at $r(0) = d(0) + 50j_r. $
$d(t) = 0 + 3j_dt.$
$r(t) = 50j_r + 4j_rt.$
Solve $d(t) = r(t) $ for t.
How far did the dog run to catch up with the rabbit?
How many leaps did the dog make?
But alas, at the last moment, the rabbit ran into a rabbit hole.
The dog ran after the rabbit into the hole and got stuck.
The rabbit ran out the rabbit hole at the other side of the borrow.
The dog had to dig himself out. The rabbit had to make repairs.
answered Mar 31 at 23:08
William ElliotWilliam Elliot
9,1812820
$endgroup$
add a comment |
$begingroup$
The rabbit jumps $j_r$ ft in a leap.
The dog jumps $j_d$ ft in a leap.
$2j_d = 3j_r $.
The dog runs at 3j_d per unit time.
The rabbit runs at 4j_r per unit time
The dog starts at d(0) = 0.
The rabbit starts at $r(0) = d(0) + 50j_r. $
$d(t) = 0 + 3j_dt.$
$r(t) = 50j_r + 4j_rt.$
Solve $d(t) = r(t) $ for t.
How far did the dog run to catch up with the rabbit?
How many leaps did the dog make?
But alas, at the last moment, the rabbit ran into a rabbit hole.
The dog ran after the rabbit into the hole and got stuck.
The rabbit ran out the rabbit hole at the other side of the borrow.
The dog had to dig himself out. The rabbit had to make repairs.
answered Mar 31 at 23:08
William ElliotWilliam Elliot
9,1812820
$endgroup$
add a comment |
$begingroup$
The rabbit jumps $j_r$ ft in a leap.
The dog jumps $j_d$ ft in a leap.
$2j_d = 3j_r $.
The dog runs at 3j_d per unit time.
The rabbit runs at 4j_r per unit time
The dog starts at d(0) = 0.
The rabbit starts at $r(0) = d(0) + 50j_r. $
$d(t) = 0 + 3j_dt.$
$r(t) = 50j_r + 4j_rt.$
Solve $d(t) = r(t) $ for t.
How far did the dog run to catch up with the rabbit?
How many leaps did the dog make?
But alas, at the last moment, the rabbit ran into a rabbit hole.
The dog ran after the rabbit into the hole and got stuck.
The rabbit ran out the rabbit hole at the other side of the borrow.
The dog had to dig himself out. The rabbit had to make repairs.
answered Mar 31 at 23:08
William ElliotWilliam Elliot
9,1812820
$endgroup$
The rabbit jumps $j_r$ ft in a leap.
The dog jumps $j_d$ ft in a leap.
$2j_d = 3j_r $.
The dog runs at 3j_d per unit time.
The rabbit runs at 4j_r per unit time
The dog starts at d(0) = 0.
The rabbit starts at $r(0) = d(0) + 50j_r. $
$d(t) = 0 + 3j_dt.$
$r(t) = 50j_r + 4j_rt.$
Solve $d(t) = r(t) $ for t.
How far did the dog run to catch up with the rabbit?
How many leaps did the dog make?
But alas, at the last moment, the rabbit ran into a rabbit hole.
The dog ran after the rabbit into the hole and got stuck.
The rabbit ran out the rabbit hole at the other side of the borrow.
The dog had to dig himself out. The rabbit had to make repairs.
answered Mar 31 at 23:08
William ElliotWilliam Elliot
9,1812820
answered Mar 31 at 23:08
William ElliotWilliam Elliot
9,1812820
answered Mar 31 at 23:08
William ElliotWilliam Elliot
9,1812820
answered Mar 31 at 23:08
William ElliotWilliam Elliot
9,1812820
9,1812820
add a comment |
add a comment |
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