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denis-greek [22]
3 years ago
12

Russell runs 9/10 mile in five minutes. How many miles does he run in one minute

Mathematics
2 answers:
shepuryov [24]3 years ago
8 0
9/50 of a mile, you have to divide 9/10 by 5 to find the number of mile(s) he ran in a minute.
dem82 [27]3 years ago
3 0
9/10=5mins
times both sides by 1/5
9/50=1min

he can run 9/50 miles per minute
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I really need help would please like help thanks
ASHA 777 [7]
Hey there!

A reciprocal might be an unfamiliar vocab word, but the concept is really quite simple.

It's just a flipped fraction! Multiplying the flipped fraction by it's reciprocal gets you one, and that's how you check your answer.

So we have:

Reciprocal of 9/10 is 10/9, because 9/10 times 10/9 = 90/90 = 1

Reciprocal of 1/11 is 11/1, because 1/11 times 11/1 = 11/11 = 1

Reciprocal of 10(remember, it's 10/1) is 1/10, because 10/1 times 1/10 = 10/10 = 1

Hope this helps!
4 0
3 years ago
Please help me solve this
andre [41]

Answer:

Step-by-step explanation:

This number is much larger than 1. You need to start at the decimal and move left (that's how scientific notation works for numbers larger than 1.)

So the answer is 1.0054 * 10 ^7

The person doing this started at the 1 and moved right until he/she hit the decimal. Wrong direction.

7 0
3 years ago
What two decimals is 79 beetween
ivolga24 [154]

Answer:

79

Step-by-step explanation:

so you dont move anything,

7 0
3 years ago
Read 2 more answers
The population of Henderson City was 3,381,000 in 1994, and is growing at an annual rate 1.8%
liq [111]
<h2>In the year 2000, population will be 3,762,979 approximately. Population will double by the year 2033.</h2>

Step-by-step explanation:

   Given that the population grows every year at the same rate( 1.8% ), we can model the population similar to a compound Interest problem.

   From 1994, every subsequent year the new population is obtained by multiplying the previous years' population by \frac{100+1.8}{100} = \frac{101.8}{100}.

   So, the population in the year t can be given by P(t)=3,381,000\textrm{x}(\frac{101.8}{100})^{(t-1994)}

   Population in the year 2000 = 3,381,000\textrm{x}(\frac{101.8}{100})^{6}=3,762,979.38

Population in year 2000 = 3,762,979

   Let us assume population doubles by year y.

2\textrm{x}(3,381,000)=(3,381,000)\textrm{x}(\frac{101.8}{100})^{(y-1994)}

log_{10}2=(y-1994)log_{10}(\frac{101.8}{100})

y-1994=\frac{log_{10}2}{log_{10}1.018}=38.8537

y≈2033

∴ By 2033, the population doubles.

4 0
3 years ago
A, B , and C are collinear points: B is between A and C. If AB = 36, BC = 5x - 9 , and AC = 54 , find x
balandron [24]

9514 1404 393

Answer:

  x = 5.4

Step-by-step explanation:

The segment sum theorem tells you ...

  AB +BC = AC

  36 +(5x -9) = 54

  5x = 27 . . . . . . . . . . . subtract 27 from both sides

  x = 5.4 . . . . . . . . . . divide by 5

4 0
3 years ago
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