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marusya05 [52]
4 years ago
5

Juan is making a fruit salad. He has grapes, watermelon, apples, pineapple, bananas, mangoes, honeydew, and cantaloupe. He wants

his fruit salad to contain five different fruits. How many ways can he make the fruit salad if it must contain watermelon? 21 35 56 70
Mathematics
2 answers:
podryga [215]4 years ago
8 0

Answer:

The correct option is B. 35

Step-by-step explanation:

Juan is making a fruit salad. He has grapes, watermelon, apples, pineapple, bananas, mangoes, honeydew, and cantaloupe

Number of fruits in total available to make fruit salad = 8

Number of fruits needed to make fruit salad = 5

No. of choices available for 1st fruit , watermelon is must = 1

\text{Total number of ways = }\frac{7!}{3!\times 4!}=35

Hence, In 35 ways he can make the fruit salad if it must contain watermelon

Therefore, The correct option is B. 35

7nadin3 [17]4 years ago
6 0
If the fruit salad must contain watermelon, he can make the fruit salad in 35 ways.
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y(x) = c_1 \textrm{e}^x + c_2\textrm{e}^{-x},

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For the second condition, we need to find the derivative y' first. In this case, we have:

y'(x) = \left(c_1\textrm{e}^x + c_2\textrm{e}^{-x}\right)' = c_1\textrm{e}^x - c_2\textrm{e}^{-x}.

Therefore:

y'(-1) = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e}^{-(-1)} = -3 \iff c_1\textrm{e}^{-1} - c_2\textrm{e} = -3.

This means that we must solve the following system of equations:

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\left(c_1\textrm{e}^{-1} + c_2\textrm{e}\right) + \left(c_1\textrm{e}^{-1} - c_2\textrm{e}  \right) = 3-3 \iff 2c_1\textrm{e}^{-1} = 0 \iff c_1 = 0.

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This means that the solution obeying the initial conditions is:

\boxed{y(x) = 3\textrm{e}^{-1} \times \textrm{e}^{-x} = 3\textrm{e}^{-x-1}}.

Indeed, we can see that:

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which do correspond to the desired initial conditions.

3 0
3 years ago
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