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natka813
1 month ago
8

the local volleyball team hosts a concession stand to raise money. They can spend $120 to purchase popcorn, candy, and drinks. t

hey purchase 95 bags of popcorn at $0.75 each and 35 bags of candy at $1.20 each. how much money does the volleyball team have left to spend on drinks?
Mathematics
1 answer:
zzz [12.3K]1 month ago
7 0
To address this inquiry, I formulated an equation:
0.75x + 1.20y = 120
0.75(95) + 1.20(35) = 120
Next, I performed the multiplication:
71.25 + 42 = 120
Now to verify:
71.25 + 42 = 113.25
The result: 120 - 113.25 = $6.75 Hope this is useful

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Answer:

1.21 g/day

Step-by-step explanation:

We start with the fact that

The mass of the bacterial colony (in grams) is described by

P(t)=2+5tan^{-1}(\frac{t}{2})

Where t=Time(in days)

Next, we differentiate with respect to t

P'(t)=5(\frac{1}{1+\frac{t^2}{4}}\times \frac{1}{2})

Using the formula \frac{d(tan^{-1}(x)}{dx}=\frac{1}{1+x^2}

P'(t)=\frac{5}{2}(\frac{4}{4+t^2})

P'(t)=\frac{10}{4+t^2}

We know that P(t)=6

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6=2+5tan^{-1}(\frac{t}{2})

5tan^{-1}(\frac{t}{2})=6-2=4

tan^}{-1}(\frac{t}{2})=\frac{4}{5}

\frac{t}{2}=tan(\frac{4}{5})

t=2tan(\frac{4}{5})

Insert the given value of t

P'(2tan\frac{4}{5})=\frac{10}{4+4tan^2(\frac{4}{5})}

P'(2tan\frac{4}{5})=\frac{10}{4}\times \frac{1}{1+tan^2(\frac{4}{5})}

We understand that 1+tan^2\theta=sec^2\theta

Applying the formula

P'(2tan(\frac{4}{5})=\frac{5}{2}\times \frac{1}{sec^2(\frac{4}{5})}

P'(2tan\frac{4}{5})=\frac{5}{2}\times cos^2(\frac{4}{5})

By employing cos^2x=\frac{1}{sec^2x}

P'(2tan\frac{4}{5})=\frac{5}{2}\times (0.696)^2=1.21g/day

Consequently, the instantaneous rate at which the mass of the colony changes is=1.21g/day

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