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3 years ago

M<2 64 and m<4 =2x-6

Mathematics

1 answer:

-Dominant- [34]3 years ago

6 0

Answer:

5iejksjsjsjsnnsnsnsnsns

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One weekday, the ticket sales person at a bus station asks every 10th ticket buyer if they had planned to use a laptop computer

andrew-mc [135]

From the 3100 weekday bus traveler, 310 travelers will be asked by the sales person. And for every 2.685 traveler questioned, a traveler will say yes. So there will be 115 bus travelers that will use a laptop computer while travelling.

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3 years ago

Please help me answer this. I’ll give a reward. To who every answers it correctly.

Nina [5.8K]

Answer:

Step-by-step explanation:

55+59 =114

180-114=66

5 0

3 years ago

Twice a number x exceed 5 at least by 4. find the possible value of x.

salantis [7]

Answer:

Step-by-step explanation:

2x › 5 (buy at least 4)

so you can just start plugging numbers in.

I used 5, 5 times 2 is 10 which is 5 more that 5, which does met the quota.

5 0

2 years ago

Read 2 more answers

Factor completely. <br> <img src="https://tex.z-dn.net/?f=x%5E%7B8%7D-%5Cfrac%7B1%7D%7B81%7D" id="TexFormula1" title="x^{8}-\fra

Eduardwww [97]

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

$x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)$

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize $x^2-\frac13$ as yet another difference of squares:

$x^2 - \dfrac13 = x^2 - \left(\dfrac1{\sqrt3}\right)^2 = \left(x-\dfrac1{\sqrt3}\right)\left(x+\dfrac1{\sqrt3}\right)$

And if you're working over the field of complex numbers, you can go even further. For instance,

$x^4 + \dfrac19 = \left(x^2\right)^2 - \left(i\dfrac13\right)^2 = \left(x^2 - i\dfrac13\right) \left(x^2 + i\dfrac13\right)$

But I think you'd be fine stopping at the first result,

$x^8 - \dfrac1{81} = \boxed{\left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)}$

6 0

3 years ago

LOOKS EASY NO FILES/LINKS THX

kompoz [17]

Answer:

D :)

Step-by-step explanation:

4 0

3 years ago