Well I need more details for x can you screenshot? It’s usually x*0.20 but I need more details to help you
Answer:
Therefore, the pairs of twin-prime numbers are (101,103) , (107,109) , (137,139) , (149,151) , (179,181) , (191,193) , (197,199) . So, the correct answer is “ (101,103) , (107,109) , (137,139) , (149,151) , (179,181) , (191,193) , (197,199) ."
2000+1200=3200=thirty two hundreds
Answer:
Step-by-step explanation:
We first need to figure out how much money Jackie has since Pilar has the same amount of money, but in different coins. Jackie has 24 dimes and 13 quarters, so the amount of money is $2.40 + $3.25 = $5.65; therefore, Pilar has $5.65 also. In order to determine how many nickels and quarters Pilar has, we have to have a system of equations. It's up to us to write the equations we need to solve this. The first equation has to do with the NUMBER of coins Pilar has, the second has to do with the VALUE of each coin. We know that Pilar has only nickels and quarters and that she has a total of 53 coins. So the NUMBER equation is
n + q = 53
The VALUE equation relates how much each coin is worth to the total amount of money it equals:
.05n + .25q = 5.65.
Go back to the first equation and solve it for either n or q. I solved it for n:
n = 53 - q
Plug this into the second equation in place of n now to get:
.05(53 - q) + .25q = 5.65 and
2.65- .05q + .25q = 5.65 and
.20q = 3 so
q = 15. This means that there are 15 quarters. If there are 15 quarters, then there are
n + 15 = 53 nickels and
n = 38 nickels.
Let's say that in the beginning he weighted x and at the end he weighted x-y, y being the number of kg he wanted to loose.
first month he lost
y/3
then he lost:
(y-y/3)/3
this is
(2/3y)/3=2/9y
explanation: ((y-y/3) is what he still needed to loose: y minus what he lost already
and then he lost
(y-2/9y-1/3y)/3+3 (the +3 is his additional 3 pounts)
(y-2/9y-1/3y)/3-3=(7/9y-3/9y)/3+3=4/27y+3
it's not just y/3 because each month he lost one third of what the needed to loose at the current time, not in totatl
and the weight at the end of the 3 months was still x-y+3, 3 pounds over his goal weight!
so: x -y/3-2/9y-4/27y-3=x-y+3
we can subtract x from both sides:
-y/3-2/9y-4/27y-3=-y+3
add everything up:
-19/27y=-y+6
which means
-19/27y=-y+6
y-6=19/27y
8/27y=6
4/27y=3
y=20.25
so... that's how much he wanted to loose, but he lost 3 less than that, so 23.25
ps. i hope I didn't make a mistake in counting, let me know if i did. In any case you know HOW to solve it now, try to do the calculations yourself to see if they're correct!
