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2 years ago

5

A bus leaves Flacq at 07 05 and reaches Curepipe at 08 50, after travelling a distance of 91 kilometres. Find the average speed

of the bus, giving your answer in kilometres per hour.

1 answer:

atroni [7]2 years ago

6 0

Time taken to travel from Flacq to Curepipe is :

8 : 50 - 7 : 05

1 hour 45 minutes = 1 + 0.75 = 1.75 hours

Distance = 91 km

Average speed is equal to :

$\boxed{ \boxed{ \frac{distance}{time} }}$

$\dfrac{91}{1.75}$

$52 \: \: kmph$

Average speed = 52 km per hour

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jeka57 [31]

Answer:

<h2> 5</h2>

Step-by-step explanation:

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Assuming there are no breaks, there are 52 weeks in a year so 475(52)= 24700

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3 years ago

5 + n2 > 8<br> A. n > 6<br> B. n>3<br> C. n>1.5<br> D. n < 26

san4es73 [151]

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3 years ago

Read 2 more answers

Please help me <br> Show your work <br> 10 points

Svet_ta [14]

<h2>Answer</h2>

After the dilation $\frac{5}{3}$ around the center of dilation (2, -2), our triangle will have coordinates:

$R'=(2,3)$

$S'=(2,-2)$

$T'=(-3,-2)$

<h2>Explanation</h2>

First, we are going to translate the center of dilation to the origin. Since the center of dilation is (2, -2) we need to move two units to the left (-2) and two units up (2) to get to the origin. Therefore, our first partial rule will be:

$(x,y)$ → $(x-2, y+2)$

Next, we are going to perform our dilation, so we are going to multiply our resulting point by the dilation factor $\frac{5}{3}$ . Therefore our second partial rule will be:

$(x,y)$ → $\frac{5}{3} (x-2,y+2)$

$(x,y)$ → $(\frac{5}{3} x-\frac{10}{3} ,\frac{5}{3} y+\frac{10}{3} )$

Now, the only thing left to create our actual rule is going back from the origin to the original center of dilation, so we need to move two units to the right (2) and two units down (-2)

$(x,y)$ → $(\frac{5}{3} x-\frac{10}{3}+2,\frac{5}{3} y+\frac{10}{3}-2)$

$(x,y)$ → $(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})$

Now that we have our rule, we just need to apply it to each point of our triangle to perform the required dilation:

$R=(2,1)$

$R'=(\frac{5}{3} x-\frac{4}{3} ,\frac{5}{3}y+ \frac{4}{3})$

$R'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(1)+ \frac{4}{3})$

$R'=(\frac{10}{3} -\frac{4}{3} ,\frac{5}{3}+ \frac{4}{3})$

$R'=(2,3)$

$S=(2,-2)$

$S'=(\frac{5}{3} (2)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})$

$S'=(\frac{10}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})$

$S'=(2,-2)$

$T=(-1,-2)$

$T'=(\frac{5}{3} (-1)-\frac{4}{3} ,\frac{5}{3}(-2)+ \frac{4}{3})$

$T'=(-\frac{5}{3} -\frac{4}{3} ,-\frac{10}{3}+ \frac{4}{3})$

$T'=(-3,-2)$

Now we can finally draw our triangle:

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3 years ago