Question

Answers in Scientific Notation.

Answer 1

one way is to factor the expressions

(a)

`10^(5)` (1.2 + (5.35 ×
`10^(1)`))

=
`10^(5)` (1.2 + 53.5)

= 54.7 ×
`10^(5)` = 5.47 ×
`10^(1)`×
`10^(5)` = 5.47 ×
`10^(6)`

(b)

`10^(-3)`((6.91 ×
`10^(1)`) + 2.4)

=
`10^(-3)`(69.1 + 2.4)

= 71.5 ×
`10^(-3)` = 7.15 ×
`10^(1)` ×
`10^(-3)` = 7.15 ×
`10^(-2)`

(c)

`10^(5)`((9.7 ×
`10^(1)`) + 8.3)

=
`10^(5)`(97 + 8.3)

= 105.3 ×
`10^(5)` = 1.053 ×
`10^(2)` ×
`10^(5)` = 1.053 ×
`10^(7)`

(d)

`10^(1)`((3.67 ×
`10^(1)`) - 1.6)

=
`10^(1)`(36.7 - 1.6)

= 35.1 ×
`10^(1)` = 3.51 ×
`10^(1)` ×
`10^(1)` = 3.51 ×
`10^(2)`

(e)

`10^(-6)`((8.41 ×
`10^(1)`) - 7.9)

=
`10^(-6)`(84.1 - 7.9 )

= 76.2 ×
`10^(-6)` = 7.62 ×
`10^(1)` ×
`10^(-6)` = 7.62 ×
`10^(-5)`

(f)

`10^(4)`((1.33 ×
`10^(1)`) - 4.9 )

=
`10^(4)`(13.3 - 4.9) = 8.4 ×
`10^(4)`