My pigg y bank has only pennies and nicklels in it , and 2/7 of the coins are nickel If i remove 84 pennies then 1/3 of the remening are pennies 1} suppose there are initially p pennies and n nickels in the pigy bank write an equation using the fact that 2/7 of the coins initially in the

There are 105 pennies in the piggy bank.

There are 42 nickels in the piggy bank.

Step-by-step explanation:

there are:

p pennies in the piggy bank

n nickels in the piggy bank

Then we can define T, the total number of coins, as:

T = p + n

We know that 2/7 of the total number of coins are nickels.

This can be written as:

n = (2/7)*T = (2/7)*(n + p)

And if we remove 84 pennies, 1/3 of the remaining coins are pennies.

This can be written as:

p - 84 = (1/3)*(n + p - 84)

Then we have a system of two equations:

n = (2/7)*(n + p)

p - 84 = (1/3)*(n + p - 84)

Let's solve the system, to do it, we first need to isolate one of the variables in one of the equations.

We can isolate n in the first one, to get:

n = (2/7)*(n + p) = (2/7)*n + (2/7)*p

n - (2/7)*n = (2/7)*p

n*(5/7) = (2/7)*p

n = (7/5)*(2/7)*p = (2/5)*p

n = (2/5)*p

Now we can replace this in the other equation:

p - 84 = (1/3)*(n + p - 84)

p - 84 = (1/3)*( (2/5)*p + p - 84)

Let's solve this for p

p - 84 = (1/3)*( (7/5)*p - 84)

3*(p - 84) = (7/5)*p - 84

3p - 252 = (7/5)*p - 84

3*p - (7/5)*p = 252 - 84

(15/5)*p - (7/5)*p = 168

(8/5)*p = 168

p = (5/8)*168 = 105

There are 105 pennies in the piggy bank.

And we know that:

n = (2/5)*p = (2/5)*105 = 42

There are 42 nickels in the piggy bank.

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)