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masya89 [10]
3 years ago
13

Maria runs 10 miles every day. If she doubles her usual speed, she can run the 10 miles in one hour less than her usual time. Wh

at is her usual speed?
Mathematics
1 answer:
Musya8 [376]3 years ago
4 0

Answer:

Her usual speed is 5 miles an hour.

Step-by-step explanation:

Since it takes one hour for her to run ten miles that means she is running at 10 miles an hour. So if you divide that by 2 it would be five.

<h2><em><u>Have a nice and great day.</u></em></h2>

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At the Olympic Games, a runner won the 26.2 mile marathon race in 2 hr 4 min and 1 second. What was his average speed in mph and
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The average speed of the runner is 12.7 mph and 20.4 km/h

Given that the runner ran 26.2 mile in 2hr and 4 minutes, we start of by converting the time from  hours and minutes into minutes and finally hours, since hours is what we need. So, we have

2hr = 120mins

+ 4 mins = 124 mins

124 mins ÷ 60 hour/mins = 2.06 hours.

This means that the runner finished the race in 2.06 hours.

If we are to find the average speed in mile per hour, we have

Average speed = distance ran ÷ time taken

Average speed = 26.2 ÷ 2.06

Average speed = 12.7 mph

From the speed in mph, we can directly convert it to km/hr by saying

1 mph = 1.609 km/h

12.7 mph = 12.7 * 1.609 = 20.4 km/hr

for more, check: brainly.com/question/1989219

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3 years ago
Whats the slope in y=−5−12x.?
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Answer:

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Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Find a compact form for generating functions of the sequence 1, 8,27,... , k^3
pantera1 [17]

This sequence has generating function

F(x)=\displaystyle\sum_{k\ge0}k^3x^k

(if we include k=0 for a moment)

Recall that for |x|, we have

\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k

Take the derivative to get

\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k

\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k

Take the derivative again:

\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k

\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k

Take the derivative one more time:

\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k

\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k

so we have

\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}

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