Amber deposited $10,000 with a bank and a five year certificate of deposit yielding 4% compound daily find the compound amount

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Ответ:

misscheoneyo

14.06.2021

Mathematics

$3^2.(3^3)^2.3^{-8} = 1$

$\frac{(3^2)(2.3)^{-3}}{2^{-3}} = \frac{1}{3}$

$\frac{(2^{-1})(3.2)^4}{(3.2)^3} = 3$

$2^5.3^5.6^{-5} =1$

$\frac{(2^3)(2.3)^{-1}}{2^2} = \frac{1}{3}$

step-by-step explanation:

the first expression is:

$3^2.(3^3)^2.3^{-8}\\=3^2.3^{3*2}.3^{-8}\\as\ the\ base\ is\ same\\=3^{(2+6-8)}\\=3^0\\=1$

second expression:

$\frac{(3^2)(2.3)^{-3}}{2^{-3}} \\= \frac{(3^2)(2^{-3}.3^{-3})}{2^{-3}}\\= \frac{3^2.2^{-3}.3^{-3}}{2^{-3}}\\=3^{2-3}\\=3^{-1}\\=\frac{1}{3}$

third expression:

$\frac{(2^{-1})(3.2)^4}{(3.2)^3} \\=\frac{(2^{-1})(3^4.2^4)}{3^3.2^3}\\=\frac{2^{(-1+4)}.3^4}{3^3.2^3}\\=\frac{2^3.3^4}{3^3.2^3}\\=\frac{3^4}{3^3}\\=3^{4-3}\\=3$

fourth expression:

$2^5.3^5.6^{-5}\\=2^5.3^5.(3.2)^-5\\=2^5.3^5.3^{-5}.2^{-5}\\=2^{(5-5)}.3^{(5-5)}\\=2^0 * 3^0\\=1*1\\=1$

fifth expression:

$\frac{(2^3)(2.3)^{-1}}{2^2}\\ =\frac{2^3.2^{-1}.3^{-1}}{2^2}\\= 2.2^{-1}.3^{-1}\\= 2^{1-1}.3^{-1}\\=2^0 * \frac{1}{3^1}\\ =1*\frac{1}{3} \\=\frac{1}{3}$ ..

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