### Question 15

### 4 comments

###### Goodness Barak

1 year agoEXPANDING .....

[(P^-8 Q⁴)/(5P^-2 Q⁴)] × [(P^6 /Q^-9)]

Q⁴ CAN NOW CANCEL EACH OTHER

[(1/5)(P^-8/P^-2)] × [P^6/Q^-9]

USING LAW OF INDICES I CAN NOW SUBTRACT THE POWERS

[(1/5)(P^-8--2)] × [P^6/Q^-9]

[(1/5)(P^-6)] × [P^6/Q^-9]

[(P^-6)/5] × (P^6/Q^-9)

MULTIPLYING....

[ P^-6 × P^6 ] /[5 ×Q^-9]

USING LAW OF INDICES I CAN NOW ADD THE POWERS

[P^-6+6]/[5Q^-9]

[P^O]/[5Q^-9]

ANYTHING RAISED TO POWER OF ZERO IS 1

1 /5Q^-9

(5Q^-9)^-1

(5^-1)(Q^-9×-1)

5^-1Q^9

(1/5) ×Q^9

(Q^9)/5.

###### Goodness Barak

1 year agoEXPANDING .....

[(P^-8 Q⁴)/(5P^-2 Q⁴)] × [(P^6 /Q^-9)]

Q⁴ CAN NOW CANCEL EACH OTHER

[(1/5)(P^-8/P^-2)] × [P^6/Q^-8]

USING LAW OF INDICES I CAN NOW SUBTRACT THE POWERS

[(1/5)(P^-8--2)] × [P^6/Q^-9]

[(1/5)(P^-6)] × [P^6/Q^-9]

[(P^-6)/5] × (P^6/Q^-9)

MULTIPLYING....

[ P^-6 × P^6 ] /[5 ×Q^-9]

USING LAW OF INDICES I CAN NOW ADD THE POWERS

[P^-6+6]/[5Q^-9]

[P^O]/[5Q^-9]

ANYTHING RAISED TO POWER OF ZERO IS 1

1 /5Q^-9

(5Q^-9)^-1

(5^-1)(Q^-9×-1)

5^-1Q^9

(1/5) ×Q^9

(Q^9)/5.