Question

Someone please help me w this standard form question​

Answer 1

`R = \frac{ {x}^(2) }{ y} \\ \\ R = \frac{ {(3.8 * {10}^(5))}^(2) }{5.9 * {10}^(4) } \\ \\ R = \frac{ ({3.8})^(2) * ( { {10}^(5) )}^(2) }{5.9 * {10}^(4) } \\ \\ R = \frac{14.44 * {10}^(10) }{5.9 * {10}^(4) } \\ \\ R = \frac{14.44 * {10}^(10 - 4) }{5.9} \\ \\ R = 2.45 * {10}^(6)`

Answer 2

2.4 × 10⁶

Explanation:

Given:

• 
  `\sf x=3.8 * 10^5`

• 
  `\sf y=5.9 * 10^4`

To calculate the value of R, substitute the given values into the given formula:

`\begin{aligned}\sf R & = \sf (x^2)/(y)\\\\ \implies \sf R& = \sf ((3.8 * 10^5)^2)/(5.9 * 10^4)\\\\& = \sf (3.8^2 * (10^5)^2)/(5.9 * 10^4)\\\\& = \sf (14.44 * 10^(10))/(5.9 * 10^4)\\\\& = \sf (14.44)/(5.9) * (10^(10))/(10^4)\\\\& = \sf 2.4474... * 10^((10-4))\\\\& = \sf 2.4474... * 10^6\\\\\end{aligned}`

Therefore, the value of R to 1 decimal place is 2.4 × 10⁶

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Exponent Rules used

`(a^b)^c=a^(bc)`

`{a^b}{a^c}=a^(b-c)`