Answer:
<1 <6 and <4 because they are on the outside of the triangle!
Step-by-step explanation:
Answer:

Step-by-step explanation:
<u>I will try to give as many details as possible. </u>
First of all, I just would like to say:

Texting in Latex is much more clear and depending on the question, just writing down without it may be confusing or ambiguous. Be together with Latex! (*^U^)人(≧V≦*)/

Note that

The denominator can't be 0 because it would be undefined.
So, we can solve the expression inside both parentheses.

Also,


Note





Note



Once


And

We have

Also, once


As



Answer:
liu;;ui;ui;iu;iu;
Step-by-step explanation:
iu;ui;iu;iiu;
That's a weird question.. i'm not sure whether it's asking for the amount of full movies watched or not, so if it means watching one the full way through, only one. if it means how much did she watch in general, than 1 1/6, or 7/6 in improper form. hopefully this helped!
Answer:
8%
Step-by-step explanation:
70.20−65.00=5.20
(5.20/65.00)×100=8