Mixed format: $$E _ { L I N E } = \sqrt { E_ R ^ { 2 } + E _X ^ { 2 } }$$ (with a space between the whole number and the numerator) Fraction format: $1 /4$
Reduce all fractions. The diameter of a steel shaft, through wear, is reduced $4 /1,000$
inch. The shaft measured $875 / 1,000$ inch originally. The electrician removes another $8 /1,000$
inch and makes a new bearing to fit the shaft. What is the new size of the shaft?

Question

Mixed format: $$E _ { L I N E } = \sqrt { E_ R ^ { 2 } + E _X ^ { 2 } }$$ (with a space between the whole number and the numerator) Fraction format: $1 /4$
 Reduce all fractions. The diameter of a steel shaft, through wear, is reduced $4 /1,000$
 inch. The shaft measured $875 / 1,000$ inch originally. The electrician removes another $8 /1,000$
 inch and makes a new bearing to fit the shaft. What is the new size of the shaft?

Quizplus · Accepted Answer

The Answer of Mixed format: $$E _ { L I N E } = \sqrt { E_ R ^ { 2 } + E _X ^ { 2 } }$$ (with a space between the whole number and the numerator) Fraction format: $1 /4$
 Reduce all fractions. The diameter of a steel shaft, through wear, is reduced $4 /1,000$
 inch. The shaft measured $875 / 1,000$ inch originally. The electrician removes another $8 /1,000$
 inch and makes a new bearing to fit the shaft. What is the new size of the shaft?

Mixed Format ELINE=ER2+EX2E _ { L I N E } = \sqrt { E_ R ^ { 2 } + E _X ^ { 2 } }ELINE​=ER2​+EX2​​

Mixed Format $E _ { L I N E } = \sqrt { E_ R ^ { 2 } + E _X ^ { 2 } }$