on a number libe point a is located at -10 and point b is located at 7 find the point on the number line that is 7/9 of the distance from a to b

Answer 1

First, we should find the distance from A to B, because we are told to find the point 7/9 of the distance from A to B.

Point A is located at -10, so it is 10 units from 0.

Point B is located at 7, so it is 7 units from 0.

Therefore the distance from A to B is 10 + 7, which is 17.

We need to find the point 7/9 of the distance from A to B, so let's turn that problem into an expression:

(location of A) + (7/9)(distance from A to B)

We add the (7/9)(distance from A to B) to the (location of A) because B is positive, and we need to follow the line to B, which is positive.

We know the location of A, which is -10, and we know the distance from A to B, which is 17, so we can substitute these values into our expression: (location of A) + (7/9)(distance from A to B)

(-10) + (7/9)(17)

Now, we multiply the second part of our expression:

(-10) + (7/9)(17) = -10 + (119/9)

Now, to add -10 and (119/9), we need to find a least common denominator. Since -10 is an integer, it is the same as (-10/1). The least common multiple of 1 and 9 is 9, so our least common denominator is 9. To keep the value of (-10/1), but change the denominator to 9, we multiply (-10/1) by (9/9) because (9/9) = 1 and multiplying something by 1 equals that something.

-10 + (119/9) = (-10/1) + (119/9) = (-90/9) + (119/9)

Now, we add (-90/9) to (119/9)

(-90/9) + (119/9) = ((-90 + 119)/9) = (29/9)

The point that is 7/9 of the distance from A to B is (29/9), which is around 3.22.

Answer 2

a and b are 17 units apart

7/9 of 17 = 119/9, or 13 2/9

-10 + 13 2/9 = 3 2/9