Answer:
Like radicals are radicals that have the same number in the radical sign
So it would be 3√7 and 11√7 (reasoning is they both have a 7 underneath the radical sign)
For the second one Simplify: 35√11-√11
you want to subtract like normal 35√11-1√11, (since there isn't a number there we are going to put one)
35-1=34
34√11, You want to keep the √11, because it is like having like terms but instead of variables it is √
34√11
The third one: Simplify: 15√6+3√6
we want to do the same thing as the problem above so,
15+3=18
18√6, again you want to keep the radical the same
18√6
For the fourth one: 5√7+√7-2√7
you want to do one step at a time so
5√7+√7= 6√7 (again you would have a one in front of the √7, then you would keep your radicals the same)
Then you want to subtract that to your other one
6√7-2√7= 4√7
6-2=4, (again keep the radical the same)
4√7
For the last one
3√5*√10+11√5*√10-√5
You always want to multiply first as in PEMDAS
Lets take this one step at a time also
First 3√5*√10
When multiplying radicals you would multiply like normal
3√50 (√5*√10= √50)
3√50
Now lets do 11√5*√10
again √5*√10=√50
so 11√50
Now you are going to add your two answers together
3√50+11√50= 14√50 (you would add 3+11=14, keep the radicals the same)
Don't forget about your -√5
14√50-√5, this as simplified as you can get so your answer is
14√50-√5
I hope this helps you ;)
Step-by-step explanation: