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V125BC [204]
3 years ago
15

How many miles did Byron drive? Please do not guess!

Mathematics
2 answers:
DaniilM [7]3 years ago
7 0

Answer:

Byron drove 10 miles. Hope this helps!

Triss [41]3 years ago
6 0
<h3>Answer:  10 miles</h3>

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

Explanation:

The distance from his house to the gas station is 5 miles. We can find this by counting out the number of spaces between the two points (since the chart indicates that 1 unit = 1 mile). Or we can subtract the x coordinates to get 8-3 = 5. This only works because the y coordinates of each point are the same value.

To find the distance from the gas station to the market, we subtract the y coordinates this time. This only works due to the x coordinates being the same. We get a distance of 8-3 = 5 miles.

Overall, the total distance Byron drives is 5+5 = 10 miles.

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Answer: if there had been no current, the rate would have been 4 km/hr.

The rate of the current is 1.5km/hr

Step-by-step explanation:

Let x represent the speed of the boat in still water.

Let y represent the speed of the current.

Impeded by the current the outing club took 4 hours and 24 minutes(4 + 24/60 = 4.4 hours) to paddle 11 km up the exeter river to their campsite last weekend. This means that the total speed with which they travelled is (x - y) km/h

Distance = speed × time. Therefore

11 = 4.4(x - y)

Dividing through by 4.4, it becomes

2.5 = x - y - - - - - - - - - - -1

The next day the same current was with them and it took only 2 hours to make the return trip to campus. This means that the total speed with which they travelled is (x + y) km/h. The distance travelled is

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Dividing through by 2, it becomes

5.5 = x + y - - - - - - - - - - -2

Adding equation 1 to equation 2, it becomes

8 = 2x

x = 8/2 = 4

Substituting x = 4 into equation 1, it becomes

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Step-by-step explanation:

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Integrate xe^(-x^2).
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You can easily integrate it using a simple substitution:

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\large\begin{array}{l} \textsf{Let}\\\\ \mathsf{-x^2=u,}\quad\textsf{then}\quad\mathsf{-2x\,dx=du}\\\\\\ \textsf{so (i) becomes}\\\\ =\mathsf{\displaystyle-\,\frac{1}{2}\int\!e^u\,du}\\\\ =\mathsf{\displaystyle-\,\frac{1}{2}\cdot e^u+C}\\\\ =\mathsf{\displaystyle-\,\frac{1}{2}\cdot e^{-2x}+C}\\\\\\ \boxed{\begin{array}{c} \mathsf{\displaystyle\int\!x\,e^{-x^2}\,dx=-\,\frac{1}{2}\,e^{-x^2}+C} \end{array}}\qquad\checkmark \end{array}


If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2159730


\large\textsf{I hope it helps. :-)}


Tags: <em>integrate substitution indefinite integral exponential composite calculus</em>

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4 years ago
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