Question

Someone please help me w this standard form question​

Answer 1

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

Answer 2

Answer:

2.4 × 10⁶

Step-by-step explanation:

Given:

• $\sf x=3.8 \times 10^5$

• $\sf y=5.9 \times 10^4$

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

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

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

--------------------------------------------------------------------------

Exponent Rules used

$(a^b)^c=a^{bc}$

${a^b}{a^c}=a^{b-c}$