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olya-2409 [2.1K]
4 years ago
12

John is packing books into boxes to send to his cousins in case they are snowed in and want to do some light reading. Each box c

an hold either 15 small books or 8 large books. He needs to pack at least 35 boxes and at least 350 books. Write a system of linear inequalities to represent the situation
Mathematics
1 answer:
Tamiku [17]4 years ago
7 0

Answer:

x+y \geq 350   ... I

\frac{x}{15}+ \frac{y}{8} \geq 35  ... II

Step-by-step explanation:

Given that John is packing books into boxes to send to his cousins in case they are snowed in and want to do some light reading.

Let x represent the number of small books and y the number of large books.

The box capacity is  either 15 small books or 8 large books

i.e. we have 15x = 8y

No of books is atleast 350 and no of boxes he needs is atleast 35

Total no of books =x+y \geq 350   ... I

is the first inequality

Next is he wants 35 boxes.

One box can hold either 15 small books or 8 large books.

For x books no of boxes required = x/15 and for y book no of boxes is y/8

Total no of boxes is atleast 35

\frac{x}{15}+ \frac{y}{8} \geq 35  ... II

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Then with the power and chain rules,

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}

Simplify this a bit by factoring out \frac12 (36-x^2)^{-3/2} :

\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}

For the second term, recall that

\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}

Then by the chain rule,

\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}

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