Answer: x/2-y^3=1 & x^2+y=4
Step-by-step explanation:
Answer:
hay 10 billetes de 2000
hay 4 billetes de 5000
Step-by-step explanation:
Planteamos las siguientes ecuaciones...:
x= billetes de 2000
y= billetes de 5000
ecuaciones :
1) x+y = 14
2) 2000x + 5000y =40000
despejamos primera ecuacion y remplazamos en segunda :
1) x= 14-y
2) 2000(14-y) + 5000y = 40000
28000- 2000y+5000y = 40000
3000y= 40000 - 28000
y= 4
reemplazamos este valor en primera ecuacion:
x+y = 14
x+ 4= 14
x= 10
ahora reemplazamo valor de X y valor de Y
2000(10) + 5000(4) =40000
40000 =40000
hay 10 billetes de 2000
hay 4 billetes de 5000
Espero que esto te ayude
Part A).
If 100 fish all had the same weight, and they weighed
235 pounds when they were all on the scale at the
same time, then each fish weighed
235 / 100 pounds .
Part-B).
You can find that actual weight with your calculator.
235 / 100 = 2.35 pounds .
Part-C).
If there were only 10 fish but their total weight was still 235 pounds,
then each fish would have to weigh a lot more, to make up the same
weight with only 10 of them.
Now, each fish would have to weigh 235 / 10 pounds .
That number is 235/10 = 23.5 pounds.
If there were only 1/10 as many fish, then in order to add up to the
same total weight, each fish would have to be 10 times as heavy.
I think :
(-4) + (-4) = -8
or
(-2) + (-2) + (-2) + (-2) = -8
but i am not sure
