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Mathematics

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8_murik_8 [283]3 years ago

6 0

hehehehehehehe.......

grin007 [14]3 years ago

3 0

$8=2^3\\4=2^2\\\\8^3\div4^3\times2^5=(2^3)^3\div(2^2)^3\times2^5\\\\\text{Use}\ (a^n)^m=a^{n\cdot m}\\\\=2^{3\cdot3}\div2^{2\cdot3}\times2^5=2^9\div2^6\times2^5\\\\\text{Use}\ a^n\times a^m=a^{n+m}\ \text{and}\ a^n\div a^m=a^{n-m}\\\\=2^{9-6+5}=2^8$

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PLEASE HELP!!!! WILL GIVE BRAINLIEST!!!In two or more complete sentences explain how you would set up and solve the following eq

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Peter's account is 910-40x, and Marla's account is 470-2x. This is a system, so you must find the number that makes both equations end up with the same number. Basically trial and error. Here is what I did:

Let's try the number 10. 910-400=510 and 470-20=450. I need a higher number.

Let's try 13 next. 910-520=390 and 470-26=444. Now the number has to be lower.

Let's try 12. 910-480=430 and 470-24=446. Close, and a little lower.

11.5 is too low, and 11.7 is too high. 11.6 has Equation 1 at 446 and Equation 2 at 446.8. Very close!

It is somewhere around 11.6. I hope that this can give you a start in figuring it out, because I don't believe in giving the complete answer, because then you do not learn anything from it.

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