If Hannah gives her younger sister 3 shirts, it does not matter what order she hands them to her. No matter the order, it will still be the same group of 3 shirt. Since order is not important this problem can be solved using a combination.
Specifically we are asked to find 8C3 (sometimes called "8 choose 3"). This is a fraction. In the numerator we start with 8 and count down 3 numbers. In the denominator we start with 3 and count all the way down to 1. Thus we obtain,
Yes I sure did hang the picture what about it
Answer:
36°
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is proved
<h3><u>
Solution:</u></h3>
Given that,
------- (1)
First we will simplify the LHS and then compare it with RHS
------ (2)

Substitute this in eqn (2)

On simplification we get,


Cancelling the common terms (sinx + cosx)

We know secant is inverse of cosine

Thus L.H.S = R.H.S
Hence proved
Answer:
See explanation
Step-by-step explanation:
a) To prove that DEFG is a rhombus, it is sufficient to prove that:
- All the sides of the rhombus are congruent:

- The diagonals are perpendicular
Using the distance formula; 








Using the slope formula; 
The slope of EG is 

The slope of EG is undefined hence it is a vertical line.
The slope of DF is 

The slope of DF is zero, hence it is a horizontal line.
A horizontal line meets a vertical line at 90 degrees.
Conclusion:
Since
and
, DEFG is a rhombus
b) Using the slope formula:
The slope of DE is 

The slope of FG is 
