Adding and subtracting mixed numbers with unlike denominators Could someone help me and explain how to do this problem thank you

Step-by-step explanation: You need to find a the lowest number that the denominators can be timed by. In you're case. You can just times 5 by 2 = 10

So that both the denominators are 10. And the denominator that was once 5. The numerator of that fraction also needs to be timed by 2.

So then you have 7 18/10 - 6 15/10

Then you can convert them to to proper fractions 8 8/10 - 7 5/10

Then you can take away the fractions so it's

8 8/10 - 7 5/10 = 1 3/10

Convert both numbers into improper fractions by multiplying the whole number by the denominator and adding the numerator

44/5 - 75/10

then you make them have the same denominator (in this case it would be 10) - (44x2)/(5x2)

88/10 - 75/10

then you take away each number as you would normally (88-75)

13/10
and finally, convert
1 3/10

Elements are of the form

 (i)

(ii)

(iii)

(iv)

(v) (a,b) where a>b=

(vi) [a,b] where a>b=

Step-by-step explanation:

Given intervals are,

(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi)  [a,b] where a>b.

To show all its elements,

(i) [a,a]

Imply the set including aa from left as well as right side.

Its elements are of the form.

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if then the set  [a,a] represents singleton sets, and singleton sets are empty so is [a,a].

(ii) [a,a)

This means given interval containing a by left and exclude a by right.

Its elements are of the form.

Since there is a singleton element a of real numbers withis the set, this set is empty.

Because there is no increment so if then the set  [a,a) represents singleton sets, and singleton sets are empty so is [a.a).

(iii) (a,a]

It means the interval not taking a by left and include a by right.

Its elements are of the form.

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if then the set  (a,a] represents singleton sets, and singleton sets are empty so is (a,a].

(iv) (a,a)

Means given set excluding a by left as well as right.

Since there is a singleton element a of real numbers, this set is empty.

Its elements are of the form.

Because there is no increment so if then the set  (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).

(v) (a,b) where a>b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

e.t.c

So this set is connected and we know singletons are connected in . Hence given set is empty.

(vi) [a,b] where .

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

e.t.c

So this set is connected and we know singletons are connected in . Hence given set is empty.