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d1i1m1o1n [39]
3 years ago
6

Keisha the trainer has two solo workout plans that she offers her clients: Plan A and Plan B. Each client does either one or the

other (not both). On Wednesday there were 2 clients who did Plan A and 3 who did Plan B. On Thursday there were 6 clients who did Plan A and 5 who did Plan B. Keisha trained her Wednesday clients for a total of 6 hours and her Thursday clients for a total of 12 hours. How long does each of the workout plans last?
Mathematics
1 answer:
gregori [183]3 years ago
6 0

Answer:

Plan A last 0.75 hours or 45 minutes.

Plan B last 1.5 hours or 90 minutes.

Step-by-step explanation:

Let a be the number of hours that the plan A last, an b the number of hours of plan B. Then for the Wednesday you have:

2a+3b=6

And for the Thursday is:

6a+5b=12

Multiply the equation of Wednesday by -3:

-6a-9b=-18

Using the method of addition using this last equation and the equation of Thursday

-6a-9b=-18\\6a+5b=12\\--------\\-4b=-6\\b=\frac{-6}{-4}\\b=1.5 hours

Replacing the value of b in one of the equations

6a+5b=12\\6a+5(1.5)=12\\6a+7.5=12\\6a=12-7.5\\a=\frac{4.5}{6} \\a=0.75hours

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max2010maxim [7]

Answer:

\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0

Step-by-step explanation:

We want to write the trignometric expression:

\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)\text{ where } u>0

As an algebraic equation.

First, we can focus on the inner expression. Let θ equal the expression:

\displaystyle \theta=\sec^{-1}\left(\frac{u}{10}\right)

Take the secant of both sides:

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Simplify. Therefore:

\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0

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Yakvenalex [24]
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