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Posted

Hey MK your acting Kaito right?

Kaito is good in math so.. answer this

-4(z+y)5(z+3) if z=7 and y=9

Answer it hahaha

Oh and I see your post *Algebcraic is easy* your so confident to yourself why answer this?

-4(z+y)5(z+3) if z=7 and y=9

-4(7+9)5(7+3)

-4(16)5(10)

-4(16)50

-64(50)

-3200

Posted

Hey MK your acting Kaito right?

Kaito is good in math so.. answer this

-4(z+y)5(z+3) if z=7 and y=9

Answer it hahaha

Oh and I see your post *Algebcraic is easy* your so confident to yourself why answer this?

Another way to solve would be this way:

-4(z+y)5(z+3) if z=7 and y=9

-4(7+9)5(7+3)

-28-36(35+15)

-28-36(50)

-64(50)

-3200

It's still the same answer.

Posted

Oops I think I used the other method like this

-4(7+9)5(7+3)

-28-36+35+15

-64+50

=-14

Ah yes, but...

PEMDAS or BODMAS before Distribution.

That's how my teacher taught the class, at least...

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