By iteratively substituting, we have



and the pattern continues down to the first term
,



Recall the formulas


It follows that



Step-by-step explanation:
First factor out the negative sign from the expression and reorder the terms
That's

<u>Using trigonometric </u><u>identities</u>
That's
<h3>

</h3>
<u>Rewrite the expression</u>
That's

We have
<h3>

</h3>
<u>Rewrite the second fraction</u>
That's
<h3>

</h3>
Since they have the same denominator we can write the fraction as

Using the identity
<h3>

</h3>
<u>Rewrite the expression</u>
We have
<h3>

</h3>
<u>Using the trigonometric identity</u>
<h3>

</h3>
<u>Rewrite the expression</u>
That's
<h3>

</h3>
Which is
<h3>

</h3>
<u>Using the trigonometric identity</u>
<h3>

</h3>
Rewrite the expression
That's
<h3>

</h3>
<u>Simplify the expression using symmetry of trigonometric functions</u>
That's
<h3>

</h3>
<u>Remove the parenthesis </u>
We have the final answer as
<h2>

</h2>
As proven
Hope this helps you
A) 0.59 uses 2 sig figs
B) 100.6 uses 4 sig figs (the zeros in this case are significant)
C) 98.42 uses 4 sig figs
D) 1.045 uses 4 sig figs (the zero is between other sig figs so it's significant)
Every choice but choice A has 4 sig figs. So the answers are B, C, and D
Answer:

Step-by-step explanation:
Use Distance Formula

Substitute

Simplify

Simplify

Add

Answer:
A.) -9
Step-by-step explanation:
it is observed that triangles ABL and KML are similar triangles
hence the following ratios are valid:
LA / LK = LB/LM = AB/KM
we can see that LA/LK = LB/LM = 1/2
hence the above ratio becomes
1/2 = AB/KM (rearranging)
KM = 2AB
We are given that KM = (x + 21) and AB = (x + 15)
hence
KM = 2AB
(x + 21) = 2 (x + 15)
x + 21 = 2x + 30
x - 2x = 30 - 21
-x = 9
x = -9