<span> I am assuming you want to prove: 
csc(x)/[1 - cos(x)] = [1 + cos(x)]/sin^3(x). 
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<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). 
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<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. 
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<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 
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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>