<span> I am assuming you want to prove:
csc(x)/[1 - cos(x)] = [1 + cos(x)]/sin^3(x).
</span>
<span>If we multiply the LHS by [1 + cos(x)]/[1 + cos(x)], we get:
LHS = csc(x)/[1 - cos(x)]
= {csc(x)[1 + cos(x)]/{[1 + cos(x)][1 - cos(x)]}
= {csc(x)[1 + cos(x)]}/[1 - cos^2(x)], via difference of squares
= {csc(x)[1 + cos(x)]}/sin^2(x), since sin^2(x) = 1 - cos^2(x).
</span>
<span>Then, since csc(x) = 1/sin(x):
LHS = {csc(x)[1 + cos(x)]}/sin^2(x)
= {[1 + cos(x)]/sin(x)}/sin^2(x)
= [1 + cos(x)]/sin^3(x)
= RHS.
</span>
<span>I hope this helps! </span>
<span>Tip and Sales tax cause a restaurant bill to____.
a. increase
When estimating using money for a purchase, you should_____.
c. estimate up to the nearest dollar or half dollar
You purchase a piece of cake for $3.49. What would be a good estimation for just purchasing the cake?
b. 3.50
Your total restaurant bill for food, drinks, and tip is $37.40. What is a good estimate cost for just the food and drinks?
c. 36.00
</span>
X=3/5
do you need an explanation?
Once the cake is removed from the oven, it has been exposed to the room temperature of 70°F. After 10 minutes, the cake's temperature decreased to 200°F, which is 150°F cooler than it initially was (350°F). As for the what is asked, 90°F is 260°F cooler than its original temperature (350°F).
This problem can be expressed in a ratio:
10mins:150<span>°F = N:260</span><span>°F
where N is how long it will take for</span><span> cake to cool to 90°F
</span>150<span>°FxN=10x260
</span>150N=2600
N=2600/150
N=17.33
N<span>≈20 minutes
Thus, it takes approximately 20 minutes </span><span>for the cake to cool to 90°F</span>
