Answer:
see explanation
Step-by-step explanation:
Using the trigonometric identity
• 1 + cot²x = csc²x
Consider the left side
x - 
x
Factorise as a difference of squares
= (csc²x - cot²x)(csc²x + cot²x)
= (1 + cot²x - cot²x)(csc²x + cot²x)
= csc²x + cot²x
= csc²x + csc²x - 1
= 2csc²x - 1 = right side ⇒ verified
Answer:
1
8
4
−
3
3
−
2
3
+
2
−
1
0
+
6
18
x
4
−
3
x
3
−
2
x
3
+
x
2
−
10
+
6
18x4−3x3−2x3+x2−10+6
1
8
4
−
3
3
−
2
3
+
2
−
4
18
x
4
−
3
x
3
−
2
x
3
+
x
2
−
4
18x4−3x3−2x3+x2−4
2
Combine like terms
Solution
1
8
4
−
5
3
+
2
−
4
Step-by-step explanation:
In order for two triangles to be similar, there must be 2 pairs of congruent angles. Since two angles of triangle PQR have different measures from two angles of triangle RST, it is impossible to have two pairs of congruent angles, so the triangles cannot be similar.
The answer is in bold type below.
<span>All
interior angles must be given to determine similarity.
The triangles
are not similar.
Similarity cannot be determined from the given
information.
The triangles are similar.
</span>
If 1/6 of the pan was left and there were 2 pieces left you just need to set 1/6 equal to 2/some number, this allows you to say that 1/6=2/x --> x/6=2-->x=12.
My explanation might be a little poor since this becomes fairly fundamental later on but hopefully it helps.
