1. I have 5 times some number of shoes. I give 10 away and now have 40. How many shoes did I originally start with?
2. I have 3 times some number of apples. I get 4 more apples. What amount of apples did I originally<span> start with? </span>
Step-by-step explanation:
cos 0=5/13. B/H
p=√h2 - b2
= √13 square -5 square
=√169 -25
=√144
=12 Ans
Tan0=p/b =12/5. , sin0=p/h =12/13,. sec0= h/b =13/5 ,. Co sec0 =h/p =13/12 , cot0 =b/p=5/12
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Answer:
y=a/(9+3x)
Step-by-step explanation:
Reverse: 9y+3xy=a
y(9+3x)=a
y=a/(9+3x)
Answer:
Step-by-step explanation:

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<u>Consider</u></h2>

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<u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>




So, on substituting all these values, we get




<h2>Hence,</h2>

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<h2>ADDITIONAL INFORMATION :-</h2>
Sign of Trigonometric ratios in Quadrants
- sin (90°-θ) = cos θ
- cos (90°-θ) = sin θ
- tan (90°-θ) = cot θ
- csc (90°-θ) = sec θ
- sec (90°-θ) = csc θ
- cot (90°-θ) = tan θ
- sin (90°+θ) = cos θ
- cos (90°+θ) = -sin θ
- tan (90°+θ) = -cot θ
- csc (90°+θ) = sec θ
- sec (90°+θ) = -csc θ
- cot (90°+θ) = -tan θ
- sin (180°-θ) = sin θ
- cos (180°-θ) = -cos θ
- tan (180°-θ) = -tan θ
- csc (180°-θ) = csc θ
- sec (180°-θ) = -sec θ
- cot (180°-θ) = -cot θ
- sin (180°+θ) = -sin θ
- cos (180°+θ) = -cos θ
- tan (180°+θ) = tan θ
- csc (180°+θ) = -csc θ
- sec (180°+θ) = -sec θ
- cot (180°+θ) = cot θ
- sin (270°-θ) = -cos θ
- cos (270°-θ) = -sin θ
- tan (270°-θ) = cot θ
- csc (270°-θ) = -sec θ
- sec (270°-θ) = -csc θ
- cot (270°-θ) = tan θ
- sin (270°+θ) = -cos θ
- cos (270°+θ) = sin θ
- tan (270°+θ) = -cot θ
- csc (270°+θ) = -sec θ
- sec (270°+θ) = cos θ
- cot (270°+θ) = -tan θ