Can you name the values of the given digits in the 3s in 330, how many times greater the first digit is than the second digit
Answers
2nd 3 is in the tenths place
the first 3 is 33 times greater
Final answer:
In the number 330, the first 3 is 10 times greater than the second 3. This is because the first 3 is in the hundreds place, making it 300 and the second 3 is in the tens place, making it 30.
Explanation:
In the number 330, the two 3's have different values. The first 3, situated in the hundreds place, has a value of 300. The second 3, situated at the tens place, has a value of 30.
To find out how many times greater the first digit is than the second digit, you divide the value of the first digit by the value of the second digit.
300 ÷ 30 = 10
So, the first digit is 10 times greater than the second digit.
Learn more about place value here:
#SPJ2
Related Questions
A room is 3 3/4 m long and 4 2/3 m wide. What is the area of the room?
Look at the picture
What fraction of a pound is each ingredient meat 0.35, vegetables 0.15, sauce 0.05, bread 0.05
How do you solve this problem? 6 x 2/5 =
Answers
------------------------------------------------
Answers
percent: 80%
decimal: 0.8
would be the fraton
Answers
Answer:
Rewrite
27
as
3^2⋅3
Pull terms out from under the radical.
3(3√3)
Multiply 3 by 3
9√3
The result can be shown in multiple forms.
Exact Form:
9√3
Decimal Form:
15.58845726
…
Answer:
3
Step-by-step explanation:
To find x, we first look at the other the side of the equation that tells us how to figure it out, which, in this case, is by finding the cubed root of 27.
The cubed root of 27 is found through multiplying a certain number by itself three times to give 27, which is commonly known to be 3- which is our x value.
We can check this by multiplying 3 by 3 by 3, mentally or arithmetically.
3 by 3 is 9, and 9 by the final 3 is 27, therefore, 3 is for sure the answer, as cubing 3 gives us 27.
An alternative method we could also use is trial and error. We'd figure it out by trialling each number from one to see if it could present us with 27.
E.g: trialling 2 cubed.
2 by 2 is 4, and 4 by 2 is 8 therefore, it cannot be two as the value is much lesser than 27.
We would then move on to the next number, this being three, we'd trial 3 cubed.
Answers
so the points are, from P1 to P2, namely P1P2, and from P2 to P3, namely P2P3, and from P3 back to P1, namely P3P1.
Answers