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rusak2 [61]
3 years ago
11

Explain how to graph 5 2 on the number line. A number line going from 0 to 3. There are 2 equals spaces between each number. The

number line contains 0, point A is one-half, 1, point B is 1 and one-half, point C is between point B and 2, 2, point D is 2 and one-half, 3.
Mathematics
2 answers:
NeTakaya3 years ago
6 0

Answer:

point D

Step-by-step explanation:

To graph 5/2 on the number line is to find its mixed number.5/2 is a improper fraction and and is equal to 2 and 1/2.So 5/2 is on point D.Because there are 2 whole groups in 5 and 1/2 that makes the mixed number.

Good luck on your test or problem

astraxan [27]3 years ago
3 0

Answer:

To graph 5/2 on the number line is to find its mixed number.5/2 is a improper fraction and and is equal to 2 and 1/2.So 5/2 is on point D.Because there are 2 whole groups in 5 and 1/2 that makes the mixed number.

Step-by-step explanation

the person above me was right

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Nathan flew 3,547 miles from Canada to California during the first part of his trip. He flew 2,567 miles from California to Hawa
AURORKA [14]

Difference in the number of miles Nathan flew between the first and second parts of his trip is 980 miles

<em><u>Solution:</u></em>

Given that Nathan flew 3,547 miles from Canada to California during the first part of his trip

He flew 2,567 miles from California to Hawaii during the second part of his trip

Therefore,

first part of his trip = 3547 miles

second part of his trip = 2567 miles

Difference in the number of miles Nathan flew between the first and second parts of his trip is given as:

difference = first part of his trip - second part of his trip

difference = 3547 - 2567 = 980

Therefore, the difference in number is 980 miles

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3 years ago
Is the square root of 9 a real number
Keith_Richards [23]
Is 3 because 3 times 3 equals 9
5 0
3 years ago
A rectangular lamina of uniform density is situated with opposite corners at (0,0) and (15,4). calculate its radii of gyration a
Sphinxa [80]
First, you have to find the moment of inertia along the x and y axes. Constant density is denoted as k.


I_{x}= \int\limits^15_0\int\limits^4_0 {k y^{2} } \, dx  dy= \frac{1}{3}k (15-4)^4=4880.33k

I_{y}= \int\limits^15_0\int\limits^4_0 {k x^{2} } \, dx  dy= \frac{1}{3}k (15-4)^4=4880.33k

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I hope I was able to help you. Have a good day.
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Answer:

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