Algebra
Calculus
Trigonometry
Matrix
Question
Simplify the expression
88222530s567
Evaluate
25335÷(22s5×802023)
Cancel out the common factor 5
567÷(22s5×802023)
Multiply the terms
567÷17644506s5
Multiply by the reciprocal
567×17644506s51
Multiply the terms
5×17644506s567
Solution
88222530s567
Show Solution
Find the excluded values
s=0
Evaluate
25335÷(22s5×802023)
To find the excluded values,set the denominators equal to 0
22s5×802023=0
Multiply the terms
17644506s5=0
Rewrite the expression
s5=0
Solution
s=0
Show Solution
Find the roots
s∈∅
Evaluate
25335÷(22s5×802023)
To find the roots of the expression,set the expression equal to 0
25335÷(22s5×802023)=0
Find the domain
More Steps
Evaluate
22s5×802023=0
Multiply the terms
17644506s5=0
Rewrite the expression
s5=0
The only way a power can not be 0 is when the base not equals 0
s=0
25335÷(22s5×802023)=0,s=0
Calculate
25335÷(22s5×802023)=0
Cancel out the common factor 5
567÷(22s5×802023)=0
Multiply the terms
567÷17644506s5=0
Divide the terms
More Steps
Evaluate
567÷17644506s5
Multiply by the reciprocal
567×17644506s51
Multiply the terms
5×17644506s567
Multiply the terms
88222530s567
88222530s567=0
Cross multiply
67=88222530s5×0
Simplify the equation
67=0
Solution
s∈∅
Show Solution