Represent the irrational number, sqrt(7) on the numberline. In order to do this, we need to know what number sqrt(7) is! It's not a nice number either, but we can figure it out. Some perfect square number that can help us 0 * 0 = 0 1 * 1 = 1 2 * 2 = 4 3 * 3 = 9 4 * 4 = 16 sqrt(7) is asking, what number times itself is 7? Let's use our perfect square numbers to help. ? * ? = 7 Where does this fit in on our list of perfect square numbers? 2 * 2 = 4 ? * ? = 7 3 * 3 = 9 7 lies between 4 and 9! This means that sqrt(7) is somewhere between... That's right, 2 and 3. In fact, 7 is closer to 9 (2 away) than 7 is from 4 (3 away). This means, sqrt(7) is closer to 3 than it is to 2 on the numberline. <-------0 1 2... (sqrt7).. 3 4 5 6 7---->
When ordering negative integers, the greatest of value will be the one with the lowest numeral and theone with the least value is the onewith the highest numeral. e.g. -1 is higher than -11. I find that if you are confused that using a numberline can be helpful. Hope that answers your question and good luck!
Well, isn't that an interesting little math puzzle we have here? To find two numbers that multiply together to give you -105, we can think of numbers like 15 and -7. When you multiply 15 by -7, you get -105. Just remember, math is like a beautiful painting - sometimes you just need to mix the right colors together to create something wonderful.
The halfway point between 1.5 and 1.6 on a number line can be found by calculating the average of the two numbers. Adding 1.5 and 1.6 together gives 3.1, and dividing by 2 gives 1.55. Therefore, 1.55 is the number that is halfway between 1.5 and 1.6 on a number line.
It is at zero = 0
A ruler
3x = 35 ??? 3x = 35 x = 35/3 ------------- ( about 11 and 7/10 )
In the middle of 0.4 and 0.5
make a numberline
ambot nimo lolo nimo
You ca see it no a numberline]
They are unit intervals.
beacause they both go chronological order
4 units
it is 75 Unfortunately, as in a very large number of cases, the above answer, from The WikiAnswers® Community is wrong.
1/infinity