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Mathematics

2 answers:

Gemiola [76]3 years ago

6 0

15 hope this helps you

Oduvanchick [21]3 years ago

6 0

Answer: 3

Step-by-step explanation:

10 had sprinkles and 2 did not which mean there are 12.

So would it be 3 that not have a sprinkles?

I think so.

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Solve this algebra...

Mumz [18]

Step-by-step explanation:

$\text{Use}\ \dfrac{a^n}{a^m}=a^{n-m}\ \text{and}\ (a^n)^m=a^{nm}\\\\\left(\dfrac{a^x}{a^y}\right)^{x-y}=\left(a^{x-y}\right)^{x-y}=a^{(x-y)(x-y)}=a^{(x-y)^2}\\\\\left(\dfrac{a^y}{a^z}\right)^{y-z}=\left(a^{y-z}\right)^{y-z}=a^{(y-z)(y-z)}=a^{(y-z)^2}\\\\\left(\dfrac{a^z}{a^x}\right)^{z-x}=\left(a^{z-x}\right)^{z-x}=a^{(z-x)(z-x)}=a^{(z-x)^2}\\\\\text{Use}\ a^n\cdot a^m=a^{n+m}\\\\a^{(x-y)^2}\cdot a^{(y-z)^2}\cdot a^{(z-x)^2}=a^{(x-y)^2+(y-z)^2+(z-x)^2}$

$\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\(x-y)^2=x^2-2xy+y^2\\(y-z)^2=y^2-2yz+z^2\\(z-x)^2=z^2-2zx+x^2\\\\=a^{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2}\\\\\text{combine like terms}\\\\=a^{(x^2+x^2)+(y^2+y^2)+(z^2+z^2)-2xy-2yz-2zx}\\\\=a^{2x^2+2y^2+2z^2-2xy-2yz-2zx}\qquad\text{distributive}\\\\=a^{2(x^2+y^2+z^2)-2(xy-yz-zx)}\\\\\text{From the equastion we know:}\ x^2+y^2+z^2=xy+yx+zx.\\\text{Therefore}\\\\=a^{2(x^2+y^2+z^2)-2(x^2+y^2+z^2)}=a^0=1$