Find 3/7 of 1/5 please helopp

8/35

Step-by-step explanation:

Step 1

We can't subtract two fractions with different denominators. So you need to get a common denominator. To do this, you'll multiply the denominators times each other... but the numerators have to change, too. They get multiplied by the other term's denominator.

So we multiply 3 by 5, and get 15.

Then we multiply 1 by 7, and get 7.

Next we give both terms new denominators -- 7 × 5 = 35.

So now our fractions look like this:

15

35

−

7

35

Step 2

Since our denominators match, we can subtract the numerators.

15 − 7 = 8

So the answer is:

8

35

Step 3

Last of all, we need to simplify the fraction, if possible. Can it be reduced to a simpler fraction?

To find out, we try dividing it by 2...

Nope! So now we try the next greatest prime number, 3...

Nope! So now we try the next greatest prime number, 5...

Nope! So now we try the next greatest prime number, 7...

Nope! So now we try the next greatest prime number, 11...

No good. 11 is larger than 8. So we're done reducing.

There you have it! The final answer is:

3/7

−

1/5

=

8/35

c. 4 and 1/2 pounds is the answer! : )

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.