Mr.Walden wrote the expression StartFraction p Superscript negative 5 Over q Superscript 0 EndFraction. He asked his students to write an equivalent expression of StartFraction p Superscript negative 5 Over q Superscript 0 EndFraction.
Four students wrote these expressions.
Isaac
Rosa
Bruce
Brianna
StartFraction q Superscript 0 Over p Superscript negative 5 EndFraction
p Superscript negative 5 + q Superscript 0
p Superscript 5 Baseline q Superscript 0
StartFraction 1 Over p Superscript 5 Baseline q Superscript 0 EndFraction
Which student wrote the correct equation that can be simplified to be equivalent to Mr. Walden’s equation?
Isaac
Rosa
Bruce
Brianna
The correct equation that can be simplified to be equivalent to Mr. Walden's equation is written by Rosa, which is StartFraction q Superscript 0 Over p Superscript negative 5 EndFraction.
Ramesh examined the pattern in the table.
Powers of 7
Value
7 Superscript 4
2,401
7 Superscript 3
343
7 Superscript 2
49
7 Superscript 1
7
7 Superscript 0
1
7 Superscript negative 1
StartFraction 1 Over 7 EndFraction
Ramesh says that based on the pattern 7 Superscript negative 5 = negative 16,807. Which statement explains whether Ramesh is correct?
Ramesh is correct because 7 Superscript negative 5 is equivalent to Negative 7 times (negative 7) times (negative 7) times (negative 7) times (negative 7), which has the same value as Negative 16,807.
Ramesh is correct because as the exponents decrease, the previous value is divided by 7, so 7 Superscript negative 5 = 1 divided by 7 divided by 7 divided by 7 divided by 7 divided by 7 = negative 16,807.
Ramesh is not correct because 7 Superscript negative 5 is equivalent to StartFraction 1 Over 7 Superscript 5 EndFraction, which has the same value as StartFraction 1 Over 7 Superscript 4 EndFraction divided by 7 = StartFraction 1 Over 7 cubed EndFraction = StartFraction 1 Over 343 EndFraction.
Ramesh is not correct because as the exponents decrease, the previous value is divided by 7, so 7 Superscript negative 5 = 1 divided by 7 divided by 7 divided by 7 divided by 7 divided by 7 = StartFraction 1 Over 16,807 EndFraction.
Ramesh is not correct because 7 Superscript negative 5 is equivalent to StartFraction 1 Over 7 Superscript 5 EndFraction, which has the same value as StartFraction 1 Over 7 Superscript 4 EndFraction divided by 7 = StartFraction 1 Over 7 cubed EndFraction = StartFraction 1 Over 343 EndFraction.
What is the pattern in the values as the exponents increase?
Powers of 3
Value
3 Superscript negative 1
One-third
3 Superscript 0
1
3 Superscript 1
3
3 squared
9
add 3 to the previous value
subtract 3 from the previous value
divide the previous value by 3
multiply the previous value by 3
The pattern in the values as the exponents increase in the table is to multiply the previous value by 3.
The correct equation that can be simplified to be equivalent to Mr. Walden's expression is: Isaac's expression, which is StartFraction q Superscript 0 Over p Superscript negative 5 EndFraction.
To determine which student wrote the correct equation that is equivalent to Mr. Walden's expression, we need to simplify each expression and compare it to the given expression.
Mr. Walden's expression: StartFraction p Superscript negative 5 Over q Superscript 0 EndFraction
Let's simplify each expression and see which one matches:
1. Isaac's expression: StartFraction q Superscript 0 Over p Superscript negative 5 EndFraction
To simplify this expression, we need to evaluate any exponent of 0, which equals 1, and any negative exponent, which equals the reciprocal. So, we have: StartFraction 1 Over p Superscript 5 Baseline EndFraction.
Isaac's expression can be simplified to: StartFraction 1 Over p Superscript 5 Baseline EndFraction
2. Rosa's expression: p Superscript negative 5 + q Superscript 0
To simplify this expression, we don't have any operations or simplifications to perform. So, Rosa's expression remains p Superscript negative 5 + q Superscript 0.
3. Bruce's expression: p Superscript 5 Baseline q Superscript 0
To simplify this expression, we need to evaluate any exponent of 0, which equals 1. So, Bruce's expression can be simplified to p Superscript 5 Baseline.
4. Brianna's expression: StartFraction 1 Over p Superscript 5 Baseline q Superscript 0 EndFraction
To simplify this expression, we need to evaluate any exponent of 0, which equals 1. So, Brianna's expression can be simplified to StartFraction 1 Over p Superscript 5 Baseline EndFraction.
Comparing the simplified expressions to Mr. Walden's expression, we find that Isaac's expression: StartFraction 1 Over p Superscript 5 Baseline EndFraction, is the correct equation that can be simplified to be equivalent to Mr. Walden’s expression.
Therefore, the student who wrote the correct equation is Isaac.