Question # 1
Given the expression
[tex]6^2\div \:3+\left(5+3\cdot \:2\right)-2^3[/tex]
Follow the PEMDAS order of operations
[tex]\mathrm{Calculate\:within\:parentheses}\:\left(5+3\cdot \:2\right)\::\quad 11[/tex]
[tex]=6^2\div \:3+11-2^3[/tex]
[tex]\mathrm{Calculate\:exponents}\:6^2\::\quad 36[/tex]
[tex]=36\div \:3+11-2^3[/tex]
[tex]\mathrm{Calculate\:exponents}\:2^3\::\quad 8[/tex]
[tex]=36\div \:3+11-8[/tex]
[tex]\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:36\div \:3\::\quad 12[/tex]
[tex]=12+11-8[/tex]
[tex]\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:12+11-8\::\quad 15[/tex]
[tex]=15[/tex]
Therefore,
[tex]6^2\div \:3+\left(5+3\cdot \:2\right)-2^3=15[/tex]
Question # 2
Given the expression
[tex]\frac{4+3^2-15\div 5}{\left(2^4-5\cdot \:3\right)^2}[/tex]
as
[tex]4+3^2-\frac{15}{5}[/tex]
[tex]=4+9-\frac{15}{5}[/tex] ∵[tex]3^2=9[/tex]
[tex]\mathrm{Add\:the\:numbers:}\:4+9=13[/tex]
[tex]=-\frac{15}{5}+13[/tex]
and
[tex]\left(2^4-5\cdot \:3\right)^2[/tex]
[tex]=\left(16-5\cdot \:3\right)^2[/tex] ∵ [tex]2^4=16[/tex]
[tex]\mathrm{Multiply\:the\:numbers:}\:5\cdot \:3=15[/tex]
[tex]=\left(16-15\right)^2[/tex]
[tex]\mathrm{Subtract\:the\:numbers:}\:16-15=1[/tex]
[tex]=1^2[/tex]
[tex]\mathrm{Apply\:rule}\:1^a=1[/tex]
[tex]=1[/tex]
Thus the equation [tex]\frac{4+3^2-15\div 5}{\left(2^4-5\cdot \:3\right)^2}[/tex] becomes
[tex]=\frac{-\frac{15}{5}+13}{1}[/tex]
[tex]\mathrm{Divide\:the\:numbers:}\:\frac{15}{5}=3[/tex]
[tex]=\frac{-3+13}{1}[/tex]
[tex]\mathrm{Apply\:rule}\:\frac{a}{1}=a[/tex]
[tex]=-3+13[/tex]
[tex]\mathrm{Add/Subtract\:the\:numbers:}\:-3+13=10[/tex]
[tex]=10[/tex]
Therefore,
[tex]\frac{4+3^2-\frac{15}{5}}{\left(2^4-5\cdot \:3\right)^2}=10[/tex]
Question # 3
Given the expression
[tex]ab-c^2+2b[/tex]
Putting a = 2, b = 4, and c = 1 in the expression
[tex]=\left(2\right)\left(4\right)-\left(1\right)^2+2\left(4\right)[/tex]
Follow the PEMDAS order of operations
[tex]\mathrm{Calculate\:exponents}\:\left(1\right)^2\::\quad 1[/tex]
[tex]=\left(2\right)\left(4\right)-1+2\left(4\right)[/tex]
[tex]\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:\left(2\right)\left(4\right)\::\quad 8[/tex]
[tex]=8-1+2\left(4\right)[/tex]
[tex]\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:2\left(4\right)\::\quad 8[/tex]
[tex]=8-1+8[/tex]
[tex]\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:8-1+8\::\quad 15[/tex]
[tex]=15[/tex]
Therefore,
[tex]ab-c^2+2b=\left(2\right)\left(4\right)-\left(1\right)^2+2\left(4\right)=15[/tex]
Question # 4
Given the expression
[tex]4d^3+2e\div \:f+de[/tex]
Putting d = 2, e = 3, and f = 6 in the expression
[tex]=4\left(2\right)^3+2\left(3\right)\div \:6+\left(2\right)\left(3\right)[/tex]
Follow the PEMDAS order of operations
[tex]\mathrm{Calculate\:exponents}\:\left(2\right)^3\::\quad 8[/tex]
[tex]=4\cdot \:8+2\left(3\right)\div \:6+\left(2\right)\left(3\right)[/tex]
[tex]\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:4\cdot \:8\::\quad 32[/tex]
[tex]=32+2\left(3\right)\div \:6+\left(2\right)\left(3\right)[/tex]
[tex]\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:2\left(3\right)\div \:6\::\quad 1[/tex]
[tex]=32+1+\left(2\right)\left(3\right)[/tex]
[tex]\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:\left(2\right)\left(3\right)\::\quad 6[/tex]
[tex]=32+1+6[/tex]
[tex]\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:32+1+6\::\quad 39[/tex]
[tex]=39[/tex]
Therefore,
[tex]4d^3+2e\div \:\:f+de=4\left(2\right)^3+2\left(3\right)\div \:6+\left(2\right)\left(3\right)=39[/tex]
