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Troyanec [42]
3 years ago
14

Henry wants to join a book-of-the-month club. The first club costs $40 to join and $10 per book. The second club costs $15 per b

ook and has no fee to join. How many books would need to be purchased from each club for the clubs to cost the same?
Mathematics
2 answers:
Marta_Voda [28]3 years ago
7 0

Answer:

There would have to be 2 books purchased from the first club. Then, there would have to be 4 books purchased from the second club.

Step-by-step explanation:

1st club.  (cost of club) 40$ + (cost of 2 books) 20$ = 60$.

2nd club. (cost of books) 15$ x  (how many books) 4 = 60$

arsen [322]3 years ago
4 0

Answer:

Step-by-step explanation:

For club 1 - 2 book

For club 2 - 4 books

Club 1 has a $40 fee + 2 books = $60 in total

Club 2 has a $0 fee + 4 book = $60 in total

I'm pretty sure I'm right

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Answer:

Step-by-step explanation:

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\int\limits^t_{0\,s} \, dt = -\frac{m}{b} \int\limits^{0\,\frac{m}{s} }_{600\,\frac{m}{s} } {\frac{1}{ v^{2}+\frac{m}{b}\cdot g } } \, dv

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t = -\frac{m}{2b}\cdot \left[\tan^{-1} \left(\frac{v}{\sqrt{\frac{m\cdot g}{b} } }\right) - \tan^{-1} \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } }\right)\right]

\tan^{-1} \left(\frac{v}{\sqrt{\frac{m\cdot g}{b} } }  \right)=-\frac{2\cdot b\cdot t}{m} + \tan^{-1}\left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } }  \right)

\frac{v}{\sqrt{\frac{m\cdot g}{b} } }=\tan \left[-\frac{2\cdot b\cdot t}{m} + \tan^{-1}\left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } }  \right)\right]

v = \sqrt{\frac{m\cdot g}{b} } \left [\frac{\tan \left(-\frac{2\cdot b \cdot t}{m}  \right)+ \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } }  \right)}{1 - \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } }  \right)\cdot \tan \left(-\frac{2\cdot b \cdot t}{m}  \right) }\right]

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<h3>The number of lines that can be drawn</h3>

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