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sladkih [1.3K]
3 years ago
7

You and a friend are running on treadmills. you run 0.5 mile every 3 minutes and your friend runs 2 miles every 14 minutes. you

both start and stop running at the same time and run a whole number of miles. what is the least possible number of miles you and your friend can run?
Mathematics
2 answers:
Natali5045456 [20]3 years ago
4 0
I run 1 mile every 6 minutes.
My friend runs 1 mile every 7 minutes.
Lowest Common Multiple:  LCM ( 6, 7 ) = 42 minutes.
Me : 42 : 6 = 7 miles
My friend : 42 : 7 = 6 miles. 

rosijanka [135]3 years ago
4 0
Start by finding how long it takes each to run 1 mile.

You: 0.5 mile every 3 minutes => 1 mile every 6 minutes

Your friend 2 miles every 14 minutes => 1 mile every 7 minutes

Now you have to calculate the minimum common multiple of 6 and 7.

For that you need to factor both numbers and take common and non common factors raised to their highest exponent:

6 = 3*2
7 = 7

Then the minimum common multiple is 2*3*7=42.

Now divide the time 42 by the humber of minutes each of you take to run 1 mile

you: 42 minutes / 6 minutes per mile = 7 miles
your friend 42 minutes / 7 minutes per mile = 6 miles.

You can run 7 miles and your friend 6 miles in the same time.

Answer: 7 and 6

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Help me with this ASAP please!
Natasha_Volkova [10]

Answer:

10 degrees

Step-by-step explanation:

The total area of the triangle will equal 180 degrees.

Firstly add 108 and 48 to get a total of 156.

Now add the 14 from (x+14). You will get 170.

Since the total would be 180 degrees, you subtract 170 from 180.

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4 0
3 years ago
What is the factoring tree for 27
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6 0
3 years ago
Five times the sum of a number and 27 is greater than or equal to six times the sum of that
AleksAgata [21]

The complete version of question:

<em>Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26. What is the solution of this problem.</em>

Answer:

5\left(x+27\right)\ge \:6\left(x+26\right)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-21\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-21]\end{bmatrix}

Step-by-step explanation:

As the description of the statement is:

'<em>Five times the sum of a number and 27 is greater than or equal to six times the sum of that number and 26'.</em>

<em />

As

  • <em>Five times the sum of a number and 27  </em>is written as: 5(x + 27)
  • <em>greater than or equal </em>is written as:  \geq
  • <em>six times the sum of that number and 26'  </em>is written as: 6(x + 26)

so lets combine the whole statement:

<em />5\left(x\:+\:27\right)\ge \:6\left(x\:+\:26\right)<em />

solving

<em />5\left(x\:+\:27\right)\ge \:6\left(x\:+\:26\right)<em />

<em />5x+135\ge \:6x+156<em />

<em />5x+135-135\ge \:6x+156-135<em />

<em />5x\ge \:6x+21<em />

<em />5x-6x\ge \:6x+21-6x<em />

<em />-x\ge \:21<em />

<em />\mathrm{Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)}<em />

<em />\left(-x\right)\left(-1\right)\le \:21\left(-1\right)<em />

<em />x\le \:-21<em />

Therefore,

5\left(x+27\right)\ge \:6\left(x+26\right)\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-21\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-21]\end{bmatrix}

6 0
3 years ago
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