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
5 answers
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.
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
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.
1 answer