Answer:
[tex]=-2x+14[/tex]
Step-by-step explanation:
[tex]\left(\frac{1}{2}\right)^2\cdot \:8-2x+\left(5+7\right)\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=\left(\frac{1}{2}\right)^2\cdot \:8-2x+5+7\\\left(\frac{1}{2}\right)^2\cdot \:8=2\\\left(\frac{1}{2}\right)^2\cdot \:8\\\left(\frac{1}{2}\right)^2=\frac{1}{2^2}\\\left(\frac{1}{2}\right)^2\\\mathrm{Apply\:exponent\:rule}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}\\=\frac{1^2}{2^2}\\\mathrm{Apply\:rule}\:1^a=1\\1^2=1\\=\frac{1}{2^2}\\=8\cdot \frac{1}{2^2}[/tex]
[tex]\mathrm{Multiply\:fractions}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}\\=\frac{1\cdot \:8}{2^2}\\\mathrm{Multiply\:the\:numbers:}\:1\cdot \:8=8\\=\frac{8}{2^2}\\\mathrm{Factor}\:8:\quad 2^3\\\mathrm{Factor\:}8=2^3\\=\frac{2^3}{2^2}\\\mathrm{Cancel\:}\frac{2^3}{2^2}:\quad 2\\\frac{2^3}{2^2}\\\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}\\\frac{2^3}{2^2}=2^{3-2}\\=2^{3-2}\\\mathrm{Subtract\:the\:numbers:}\:3-2=1\\=2\\=2-2x+5+7\\Group\:like\:terms\\=-2x+2+5+7[/tex]
[tex]\mathrm{Add\:the\:numbers:}\:2+5+7=14\\=-2x+14[/tex]
