Question

How do you evaluate `c-2ab` if `a=2.4, b=0.237`, and `c=9.49`?

Answer 1

`8.3524`

Explanation:

Substituting the values we have:

`c-2ab" "->" "9.49-2(2.4)(0.237)`
..........................................................................................
Multiplying out the brackets we have:

If the decimals give you a problem and you wish to calculate manually you may do this:

`2.4` is the same as `24xx1/10`
`0.237` is the same as `237xx1/1000`

Putting it all together we have:

`24xx237xx1/10xx1/1000 color(white)("d")->color(white)("d")24xx237xx1/10000`

Multiplying out just the whole numbers:
`20xx237->4740`
`color(white)("d")4xx237->ul(color(white)("4")948 larr" Add")`
`color(white)("dddddddddd")5688`
Now we multiply by `1/10000` giving: `0.5688` giving:

`9.4200-2(0.5688)`

`9.4200-1.1376`
......................................................................................

Changing the way that 9.4900 looks without changing its value

`9.4900" is the same as "9.4cancel(9)^8 cancel(0)^(10)0`

and `9.4cancel(9)^8 cancel(0)^(10) 0" is the same as "9.4(cancel(9))^8 cancel(0)^(9) cancel(0)^10`
...............................................................................

Now we can do the subtraction more easily:

`9.4cancel(9)^8 cancel(0)^(9) cancel(0)^10`
`ul(1.1color(white)(.)3color(white)(.)7color(white)(.)6)" "larr" Subtract"`
`8.3color(white)(.)5color(white)(.)2color(white)(.)4`

color(white)("d")