Hey there! We are given two decimal numbers to compare.
52.311 and 52.31
These two numbers are similar except that 52.311 has another 1 in while 51.31 doesn't.
Keep in mind that 52.31 can also refer to 52.310
So we are comparing 52.311 and 52.310
Since 1 is greater than 0.
52.311 is greater than 52.31 (52.311 > 52.31)
Let me know if you have any questions!
Topic: Decimal Numbers
Answer:
21
Step-by-step explanation:
Let x represent the number of dimes and y represent the number of nickels. The total number of coins is 37; this gives us the equation
x+y = 37
Each dime is worth ten cents, or 0.10, and each nickel is worth five cents, or 0.05. The total amount of money is given by
0.10x+0.05y = 2.65
This gives us the system
To solve this, we will use substitution. We will isolate x in the first equation:
x+y=37
Subtract y from each side:
x+y-y = 37-y
x = 37-y
Substitute this into the second equation:
0.10(37-y)+0.05y = 2.65
Using the distributive property,
0.10(37)-0.10(y)+0.05y = 2.65
3.70-0.10y+0.05y = 2.65
Combining like terms,
3.70-0.05y = 2.65
Subtract 3.70 from each side,
3.70-0.05y-3.70 = 2.65-3.70
-0.05y = -1.05
Divide both sides by -0.05:
-0.05y/-0.05 = -1.05/-0.05
y = 21
There were 21 nickels.
If the whole diameter of sphere is 6 inch(means radius is 3 inch) & thickness of material is 1 inch, so the radius of the hollow part should be 2 inch (3-1=2)
The total volume of the ball will be;
= Total volume - Volume of hollow part
= 4/3πr³ - 4/3πr³
= 4/3 × 3.14 × (3)³ - 4/3×3.14×(2)³
= 4 × 3.14 × 9 - 4/3 × 3.14 × 8
= 113.04 - 33.52
= 79.52
If the density of the material is 3.25 g/in3, then the density of 79.52 inch³ should be;
= 79.52 × 3.25
= 258.44 g
Now, if there are 8 spheres in one box, the weight if box will be;
= 258.44 × 8
= 2,067.52 g
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Ahh! Finally done with this, if you have any doubt just do ask me :)
