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3 years ago

My number is a perfect square the only prime number in its prime factorization is 2

1 answer:

8 0

I think the number would be 256 because

256 :
128*2
64*2
32*2
16*2
8*2
4*2
2*2

secondly if u divide 256 by 32 it gives 8 so it's a factor of 32

and lastly 2+5+6 = 13 which is an odd number

hope this helps!

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What is 19,067 divided by 43

Zigmanuir [339]

It could be 443, try simplifying to the nearest hundreaths, hope this helps!

8 0

2 years ago

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Write a decimal in tenths that is less than 3.81 but greater than 3.0

marin [14]

Maybe 38/10. Transforming it into fractions, then in tenths

4 0

3 years ago

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0.2.0.4, 0.6, 0.8, 1....<br> Arithmetic or geometric

Basile [38]

Arithmetic sequences have a common difference between consecutive terms.

Geometric sequences have a common ratio between consecutive terms.

Let's compute the differences and ratios between consecutive terms:

Differences:

$0.4-0.2 = 0.2,\quad 0.6-0.4=0.2,\quad 0.8-0.6=0.2,\quad 1-0.8=0.2$

Ratios:

$\dfrac{0.4}{0.2}=2,\quad \dfrac{0.6}{0.4} = 1.5,\quad \dfrac{0.8}{0.6} = 1.33\ldots, \quad\dfrac{1}{0.8}=1.25$

So, as you can see, the differences between consecutive terms are constant, whereas ratios vary.

So, this is an arithmetic sequence.

3 0

2 years ago

Can someone help me this question please?

pashok25 [27]

Larger number is 32 because if u take 48 minus 16 u get your anwser

5 0

3 years ago

Read 2 more answers

How do I solve this question?

Sati [7]

Answer:

m=2 and n=3

Step-by-step explanation:

<u>Step</u> :-

Given $[ 2 x^{n}y^{2} ]^m = 4 x^6 y^4$

using algebraic formula $(a^m)^n= a^{mn}$

now

$2^{m} x^{mn} y^{2m} = 4x^{6} y^{4}$

now equating 'x' powers, we get

$x^{mn} = x^{6}$

$mn=6$ ....(1)

now

$y^{2m} = y^{4}$

Equating 'y' powers ,we get

2 m=4

m=2

substitute m= 2 in equation (1)

we get

2 n=6

n=3

verification:-

substitute m=2 and n=3 , we get

$[ 2 x^{n}y^{2} ]^m = 4 x^6 y^4$

$(2 x^{3} y^2)^2 = 4 x^6 y ^4\\$

$4 x^6 y^4 = 4 x^6 y^4$

both are equating so m= 2 and n=3

4 0

2 years ago