What is 6500(1+(0.028)/(12))^{12*2} ?

The solution to 6500(1+(0.028)/(12))^{12*2} is (13*3007^{24})/(2^{70)*3^{24}*125^{23}}

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simplify 6500(1+(0.028)/(12))^{12*2}

simplify

6500 (1 + \frac{0.028}{12})^{12 \cdot 2}

\frac{13 \cdot 300 7 ^{24}}{2 ^{70} \cdot 3 ^{24} \cdot 12 5 ^{23}}

Decimal Notation

6873.93652 \dots

Solution steps

6500 (1 + \frac{0.028}{12})^{12 \cdot 2}

Multiply the numbers:

12 \cdot 2 = 24

= 6500 (1 + \frac{0.028}{12})^{24}

Simplify

1 + \frac{0.028}{12} : \frac{3007}{3000}

= 6500 (\frac{3007}{3000})^{24}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

(\frac{3007}{3000})^{24} = \frac{300 7 ^{24}}{300 0 ^{24}}

= 6500 \cdot \frac{300 7 ^{24}}{300 0 ^{24}}

6500 \cdot \frac{300 7 ^{24}}{300 0 ^{24}} = \frac{13 \cdot 300 7 ^{24}}{2 ^{70} \cdot 3 ^{24} \cdot 12 5 ^{23}}

= \frac{13 \cdot 300 7 ^{24}}{2 ^{70} \cdot 3 ^{24} \cdot 12 5 ^{23}}

(x^{\frac{2}{3}})^{- \frac{3}{4}}

4 (x - 2) = 12

- 5 x^{2} + 11 x + 12

\frac{x + 4}{x - 4} \leq 0

25 x^{2} - 81 = 0

What is 6500(1+(0.028)/(12))^{12*2} ?
The solution to 6500(1+(0.028)/(12))^{12*2} is (13*3007^{24})/(2^{70)*3^{24}*125^{23}}