Answer: 40G+8D ≤ 80; D must be at least 2 as stated in problem and can be up to 10, and you can have a maximum of 1 game book depending on how many dice sets you get (if you get more than 5 dice sets you cannot have any game books).
Step-by-step explanation: Since gaming books cost $40 so $40G for the amount of money spent and since sets of dice cost $8 so 8D for the money spent. Since the money spent must be less than or equal to 80 we put symbol at end.
F = 9/5 C + 32
- 32 -32
subtracting - 32 from both sides
F - 32 = 9/5c
5/9 F -32 = 5/9(9/5)C
Invert 9/5 to 5/9 multiply to both sides
5/9F - 32 = C
The answer is 5/9F-32 = C
Answer:
395 minutes
Step-by-step explanation:
See above for the answer
Answer:
Step-by-step explanation:

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

<h2>
<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
Assuming width=diameter
diameter/2=radius
area=hpir^2
w=d=2
d/2=2/2=1=r
h=6
area=6pi1^2
area=6pi1
area=6pi
aprox pi=3.141592
area=18.85 cm³