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wel
3 years ago
13

1. Is the number 5 prime, composite, or neither?

Mathematics
2 answers:
MaRussiya [10]3 years ago
8 0

Answer:

5 is a prime that's the answer

Elena-2011 [213]3 years ago
3 0
<h2>1. Is the number 5 prime, composite,</h2><h2>or neither?</h2><h3>The number 5 is prime since it can only be divided by itself and 1.</h3><h3 /><h3>Hope I helped. :)</h3>
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What is the greatest 0.45 0.097 0.375 0.209
Nady [450]

Answer:

Largest number is : 0.45

Step-by-step explanation:

0.45, 0.097, 0.375, 0.209

Here all numbers have first digit is 0.

So, look at the first decimal place from all decimal numbers.

4 from .45, 0 from .097, 3 from .375 and 2 from .209

Largest digit is 4

So, largest digit is 0.45

So, final answer is 0.45

I hope it will help you.

If you have any doubt, please leave a comment. I'll help you.

3 0
3 years ago
Read 2 more answers
) a circle with radius 63mm
Jet001 [13]

Answer:

<em>radius: r2 = 3.14 × 632 = 12470 square mm</em>

Step-by-step explanation:

5 0
3 years ago
2. Find the volume for the rectangular prism.<br> 8 m<br> 4 m<br> 7 1/4 m
BlackZzzverrR [31]
V= 232 m (8x4x7.25) 232 is the volume
5 0
3 years ago
A professor has recorded exam grades for 10 students in his​ class, but one of the grades is no longer readable. If the mean sco
Nostrana [21]

Answer:

unreadable score = 35

Step-by-step explanation:

We are trying to find the score of one exam that is no longer readable, let's give that score the name "x". we can also give the addition of the rest of 9 readable s scores the letter "R".

There are two things we know, and for which we are going to create equations containing the unknowns "x", and "R":

1) The mean score of ALL exams (including the unreadable one) is 80

so the equation to represent this statement is:

mean of ALL exams = 80

By writing the mean of ALL scores (as the total of all scores added including "x") we can re-write the equation as:

\frac{R+x}{10} =80

since the mean is the addition of all values divided the total number of exams.

in a similar way we can write what the mean of just the readable exams is:

\frac{R}{9} = 85\\ (notice that this time we don't include the grade x in the addition, and we divide by 9 instead of 10 because only 9 exams are being considered for this mean.

Based on the equation above, we can find what "R" is by multiplying both sides by 9:

\frac{R}{9} = 85\\R=85*9= 765

Therefore we can now use this value of R in the very first equation we created, and solve for "x":

\frac{R+x}{10} =80\\\frac{765+x}{10} =80\\765+x=80*10=800\\765+x=800\\x=800-765=35

4 0
3 years ago
Particle 1 of charge q1 �� ��5.00q and particle 2 of charge q2 �� ��2.00q are fixed to an x axis. (a) as a multiple of distance
Naya [18.7K]

<span>Assuming that the particle is the 3rd particle, we know that it’s location must be beyond q2; it cannot be between q1 and q2 since both fields point the similar way in the between region (due to attraction). Choosing an arbitrary value of 1 for L, we get </span>

<span>
k q1 / d^2 = - k q2 / (d-1)^2 </span>

Rearranging to calculate for d:

<span> (d-1)^2/d^2 = -q2/q1 = 0.4 </span><span>
<span> d^2-2d+1 = 0.4d^2 </span>
0.6d^2-2d+1 = 0  
d = 2.72075922005613 
d = 0.612574113277207 </span>

<span>
We pick the value that is > q2 hence,</span>

d = 2.72075922005613*L

<span>d = 2.72*L</span>

3 0
3 years ago
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