Last thursday, john assembled chairs at a rate of 3 chairs per hour for part of the day and larry assembled no chairs. last friday, larry assembled chairs at a rate of 4 chairs per hour for part of the day and john assembled no chairs. if john and larry assembled chairs for a total of 7 hours during these two days, how many chairs did john assemble last thursday?

9 chairs

Step-by-step explanation:

Let the complete question is,

"Last Thursday, John assembled chairs at a rate of 3 chairs per hour for part of the day and Larry assembled no chairs. Last Friday, Larry assembled chairs at a rate of 4 chairs per hour for part of the day and John assembled no chairs. If John and Larry assembled chairs for a total of 7 hours during these two days, how many chairs did John assemble last Thursday if  during these two days, John and Larry assembled a total of 25 chairs."

Let last Thursday, John assembled chairs for x hours and last Friday, Larry assembled chairs for y hours,

So, total number of hours = x + y

∵ total hours of working = 7,

⇒ x + y = 7,    (1)

Rate of assembling chairs for John = 3 chairs per hour,

Thus, the total chairs assembled by John = 3x,

Similarly, Rate of assembling chairs for Larry = 4 chairs per hour,

Thus, the total chairs assembled by Larry = 4y,

Total assembled chairs = 3x + 4y

∵ Total assembled chairs = 25,

⇒ 3x + 4y = 25      (2)

Equation (2) - 4 × equation (1),

3x - 4x = 25 - 28

-x = -3

⇒ x = 3

Hence,

The number of chairs assembled by John last Thursday = 3(3) = 9

13

step-by-step explanation:

distance formula is d=√(x2-x1)^2   +   (y2-y1)^2

51

step-by-step explanation:

73-22=51 the   mass number minus the amount of protons gives you the amount of neutrons