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Vera_Pavlovna [14]
3 years ago
12

Two planes leave an airport traveling in opposite directions. One plane travels at 540 kilometers per hour and the other at 520

kilometers per hour. How long will it take them to be 2650 kilometers​ apart?
Mathematics
2 answers:
Daniel [21]3 years ago
8 0
<h3>Answer:    2.5 hours</h3>

This is equivalent to 2 hours, 30 minutes

=========================================

Explanation:

x = number of hours

Let's say we have two plans A and B. Plane A travels 540 km/hr and let's make it travel eastward. Plane B would travel the opposite direction (west) at 520 km/hr.

After x hours, plane A travels 540x kilometers. Plane B travels 520x km after the same amount of time, since it takes off at the same time plane A does.

In total, the distance between the two planes after x hours is 540x+520x = 1060x kilometers.

Set this equal to 2650 and solve for x

1060x = 2650

1060/1060 = 2650/1060 ... divide both sides by 1060

x = 2.5

It takes 2.5 hours for the two planes to be 2650 km apart.

--------

side note:

2.5 hrs = 2 hrs + 0.5 hrs

2.5 hrs = 2 hrs + (0.5 hrs)*(60 min/1 hr)

2.5 hrs = 2 hrs + 30 min

2.5 hrs = 2 hrs, 30 min

Dennis_Churaev [7]3 years ago
7 0

Answer: at twice the father's rate of speed. At this rate; how many miles would the bicycle rider travel in 9 hours?

Step-by-step explanation:

at twice the father's rate of speed. At this rate; how many miles would the bicycle rider travel in 9 hours?

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Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.
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Answer:

\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}

Step-by-step explanation:

To find the derivative of the function y(x)=\ln \left(\frac{x}{x^2+1}\right) you must:

Step 1. Rewrite the logarithm:

\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }

Step 2. The derivative of a sum is the sum of derivatives:

\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }}={\left(\left(\ln{\left(x \right)}\right)^{\prime } - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }\right)

Step 3. The derivative of natural logarithm is \left(\ln{\left(x \right)}\right)^{\prime }=\frac{1}{x}

{\left(\ln{\left(x \right)}\right)^{\prime }} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }={\frac{1}{x}} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }

Step 4. The function \ln{\left(x^{2} + 1 \right)} is the composition f\left(g\left(x\right)\right) of two functions f\left(u\right)=\ln{\left(u \right)} and u=g\left(x\right)=x^{2} + 1

Step 5.  Apply the chain rule \left(f\left(g\left(x\right)\right)\right)^{\prime }=\frac{d}{du}\left(f\left(u\right)\right) \cdot \left(g\left(x\right)\right)^{\prime }

-{\left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }} + \frac{1}{x}=- {\frac{d}{du}\left(\ln{\left(u \right)}\right) \frac{d}{dx}\left(x^{2} + 1\right)} + \frac{1}{x}\\\\- {\frac{d}{du}\left(\ln{\left(u \right)}\right)} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- {\frac{1}{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}

Return to the old variable:

- \frac{1}{{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- \frac{\frac{d}{dx}\left(x^{2} + 1\right)}{{\left(x^{2} + 1\right)}} + \frac{1}{x}

The derivative of a sum is the sum of derivatives:

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Step 6. Apply the power rule \frac{d}{dx}\left(x^{n}\right)=n\cdot x^{-1+n}

\frac{1}{x^{3} + x} \left(x^{2} - x \left({\frac{d}{dx}\left(x^{2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(x^{2} - x \left({\left(2 x^{-1 + 2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x \frac{d}{dx}\left(1\right) + 1\right)\\

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Thus, \frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}

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-(c^2+4c-21)

Let’s focus on the quadratic expression inside the parentheses. To find our factors (c + x)(c + y), we’ll need to find two terms x and y that multiply together to make -21 and add together to make 4. It turns out that the numbers -3 and 7 work out perfectly for that purpose (-3 x 7 = -21 and 7 + (-3) = 4), so substituting them in for x and y, we have:

(c + (-3))(c + 7)
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And adding back on the negative from a few steps earlier:

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