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AleksandrR [38]
3 years ago
14

You wànker YOU WÀNKER

Mathematics
1 answer:
Sidana [21]3 years ago
3 0

Answer:

lol

Step-by-step explanation:

hi hi

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2 to the power 948 equals what
Natali5045456 [20]
Hello,

2^948=2,3792270535644529004768999970398e+285
=2379227053564452900476899997039840896210016322655031134489234974905505051456646997672269303193850160943677958064308756880727336392871849132465328929763831401252753344715935798308298255734876378992382713251762299529708397931004608141051358304557852932819272168726630260518024558103494656


8 0
4 years ago
Solve for x. <br><br> 3x/6 + 1 = 7
Evgesh-ka [11]
The answer for 3x/6+1=7 is 12
5 0
3 years ago
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I NEED HELP ASAP<br> solve for v<br><br> v/10.24=8
11Alexandr11 [23.1K]

Answer:

v/10.24=8 V=4 <-- I am not sure if it's right but, I hope this helps u, Srry if it's the wrong answer

Step-by-step explanation:

5 0
3 years ago
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True or false in the function.......​
malfutka [58]
It’s false. it’s +2 not -2
4 0
3 years ago
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$i^5+i^{-25}+i^{45}$<img src="https://tex.z-dn.net/?f=%24i%5E5%2Bi%5E%7B-25%7D%2Bi%5E%7B45%7D%24" id="TexFormula1" title="$i^5+i
gregori [183]

i^5+i^{-25}+i^{45}

Rewrite the term i^{-25}

=i^5+\dfrac{1}{i^{25}}+i^{45}

Expand each term so we have

=i(i^2)^2+\dfrac{1}{i(i^2)^{12}}+i(i^2)^{22}

Use the fact that i^2=-1

=i(-1)^2+\dfrac{1}{i(-1)^{12}}+i(-1)^{22}

Use the fact that (-1)^{a}=1 when a is an even number

=i+\dfrac{1}{i}+i

Simplify

=i-i+i

=i

Let me know if you need any clarifications, thanks!

3 0
3 years ago
Read 2 more answers
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