Question

determine whether each statement is true or false.
$\frac{1}{4}(2^{3}-5 + 1)=1$ select choice
$81div(4^{2}-10 + 2)=1$ select choice
$\frac{7^{2}-8(5)-1}{8}+12(\frac{1}{12})=1$ select choice

Explanation:

Step1: Simplify the left - hand side of the first equation

First, calculate the exponent and operations inside the parentheses: $2^3=8$, then $2^3 - 5+1=8 - 5 + 1=4$. Then $\frac{1}{4}(2^3 - 5 + 1)=\frac{1}{4}\times4 = 1$.

Step2: Simplify the left - hand side of the second equation

First, calculate the operations inside the parentheses: $4^2=16$, then $4^2-10 + 2=16-10 + 2=8$. Then $1\div(4^2 - 10 + 2)=1\div8=\frac{1}{8}$, and $8[1\div(4^2 - 10 + 2)]=8\times\frac{1}{8}=1$.

Step3: Simplify the left - hand side of the third equation

First, calculate the numerator of the first fraction: $7^2=49$, $8\times5 = 40$, so $7^2-8\times5-1=49-40 - 1=8$. Then $\frac{7^2-8\times5 - 1}{8}=\frac{8}{8}=1$, and $12\times\frac{1}{12}=1$. So $\frac{7^2-8\times5 - 1}{8}+12\times\frac{1}{12}=1 + 1=2
eq1$.

Answer:

The first statement is True.
The second statement is True.
The third statement is False.