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Elina [12.6K]
2 years ago
10

Round 72,934 to the nearest hundreds place and enter the value below.

Mathematics
2 answers:
Ganezh [65]2 years ago
8 0
72,900 should be the answer
anastassius [24]2 years ago
5 0

Answer:

72,900

Step-by-step explanation:

9 is in the hundreds place so therefore you should look to the left which us 3 and 3 is less than 5 so the answer is 72,900

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A collection of coins consists of nickels, dimes, and quarters. There are three fewer quarters than nickels and six more dimes t
weeeeeb [17]

Answer:

You have 15 dimes. You have 9 quarters. You have 12 nickels

Step-by-step explanation:

lets set some variables:

let "n" = the number of nickels

let "d" = the number of dimes

let "q" = the number of quarters

So, the total amount of money you have should be: $4.35 = 0.25q + 0.10d + 0.05n

Now let's look at the relationships between the coins:

"There are three fewer quarters than nickels": n - 3 = q

"six more dimes than quarters": q + 6 = d

So now you have three equations with three variables, all you need to do is solve.

\left \{ {{4.35 = 0.25q + 0.10d + 0.05n} \atop {n - 3 = q}} \atop {q + 6 = d}}\right.

first, you can substitute "n-3" for "q" (according to the 2nd equation) in the 1st and 3rd equation, you get:

\left \{ {{4.35=0.25(n-3)+0.10d+0.05n} \atop {(n-3)+6=d}} \right.

You now only have two equations and two variables.

Simplify:

\left \{ {{4.35=0.25n-0.75+0.10d+0.05n} \atop {n+3=d}} \right.

\left \{ {{4.35=0.30n-0.75+0.10d} \atop {n+3=d}} \right.

Now substitute "n+3" for "d" (according to the 2nd equation) in the 1st equation:

4.35=0.30n-0.75+0.10(n+3)

simplify:

4.35=0.30n-0.75+0.10n+0.30

4.35=0.40n-0.45

4.35+0.45=0.40n

4.80=0.40n

n=12

You have 12 nickels. Now sub "n" back into your equations to find the number of dimes and quarters:

n - 3 = q

12 - 3 = q

q = 9

You have 9 quarters.

q + 6 = d

9 + 6 = d

d = 15

You have 15 dimes.

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