The formula for depreciation is:

Where x = Initial value,
y= Amount after depreciation.
r= Rate of depreciation,
t = time (in years)
According to given problem,
x = 1040, y= 944 and t = 12 months =1 year.
So, first step is to plug in these values in the above formula, So,

944 = 1040 (1 -r)
 Divide each sides by 1040.
 Divide each sides by 1040.
0.907692308 =1 - r
0.907692308 - 1 = -r Subtract 1 from each sides.
-0.092307692 = -r
So, r = 0.09 or 9%.
Now plug in 0.09 in the above equation to get the depreciation equation. So,

So, 
b) To find the value of the bike after 5 months, 
plug in t = 5 months= 5/12 = 0.41667 years in the above equation of depreciation.
So, 
y = 1040 * 0.961465659
y = 999.9242852
y = 1000 (Rounded to nearest integer).
Hence, the value of the bike after 5 months is $1000.
 
        
             
        
        
        
Answer:
4cotα=tanα
4(1/tanα)=tanα
(4/tanα)=tanα
cross multiply 
=> 4=tan²α
√4=√tan²α
±2=tanα
α=arc( tan) |2|
α=63.4° ( in first quadrant) 
and
α=180+63.4=243.4 in the third quadrant 
since we also found a negative answer( i.e –2) then α also lies in quadrants where it gives a negative value(i.e second and fourth quadrants) 
α=180–63.4=116.6° in the second quadrant 
α=360–63.4=296.6 in the fourth quadrant 
therefore theta( in my case, alpha) lies in all four quadrants and is equal to:
α=63.4°,243.4°,116.6°and 296.6°
 
        
             
        
        
        
Find the area of the base  .
.
Multiply area of the base by height to get volume

Hope this helps.
 
        
                    
             
        
        
        
6x + 4y = 22 
+
 2x - y = 1
____________
 8x + 3y = 23
        
             
        
        
        
Answer:
- $5000 at 10%, $10000 at 12% and 10000 at 16%
Step-by-step explanation:
- <em>One part of $ 25,000 is invested at 10% interest, another part at 12%, and the rest at 16%. The total annual income from the three investments is $ 3,200. Also, the income from the investment at 16% is equal to the income from the other two investments combined. How much was invested at each interest rate?</em>
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Let the parts be x, y and z
<u>As per given we get below system of equations:</u>
- x + y + z = 25000
- 0.1x + 0.12y + 0.16z = 3200
- 0.1x + 0.2y = 0.16z
<u>Substitute 0.1x + 0.2y in the second equation:</u>
- 0.16z + 0.16z = 3200
- 0.32z = 3200
- z = 3200/0.32
- z = 10000
<u>Now we have:</u>
- x + y + 10000 = 25000 ⇒ x + y = 15000
and
- 0.1x + 0.12y + 0.16*10000 = 3200 ⇒ 0.1x + 0.12y = 1600
<u>Multiply the second equation and then subtract the first one:</u>
- 10(0.1x + 0.12y) = 10(1600) ⇒ x + 1.2y = 16000
- x + 1.2y - (x + y) = 16000 - 15000
- 0.2y = 1000
- y = 10000
Then 
<u>So the parts are:</u>
- $5000 at 10%, $10000 at 12% and 10000 at 16%