Rearrange into the form "x^2 + bx + C = 0" then solve by factorising x^2 - 21 = 4x

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kayltosh2729

15.02.2023

Mathematics

$\dfrac{a \times 10^{m}}{b \times 10^{-n}} = \dfrac{a}{b} \times 10 ^{m - (- n)} = \dfrac{a}{b} \times 10^{m+n}$ above i used exponent properties.the result will be a relatively large number, compared to the original. the exponent on the 10 would be larger than either of the original exponents.

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