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2 years ago

Find the approximate sum of the series shown below.

Mathematics

1 answer:

fomenos2 years ago

4 0

Answer:

$The~expression=7(\frac{5}{3})^{(1-1)}+ 7(\frac{5}{3})^{(2-1)}+7(\frac{5}{3})^{(3-1)}+7(\frac{5}{3})^{(4-1)}+7(\frac{5}{3})^{(5-1)}\\~~~~~~~~~~~~~~~~~~~~~~=7(\frac{5}{3})^0+7(\frac{5}{3})^1+7(\frac{5}{3})^2+7(\frac{5}{3})^3+7(\frac{5}{3})^4\\~~~~~~~~~~~~~~~~~~~~~~=7(1)+\frac{35}{3}+7(\frac{25}{9})+7(\frac{125}{27}) +7(\frac{625}{81})\\~~~~~~~~~~~~~~~~~~~~~~=7+\frac{35}{3}+\frac{175}{9}+\frac{875}{27} +\frac{4375}{81} \\$

$=\frac{7(81)}{81}+\frac{35(27)}{3(27)} +\frac{175(9)}{9(9)} +\frac{875(3)}{27(3)}+\frac{4375}{81}\\=\frac{7(81)+35(27)+175(9)+875(3)+4375}{81}\\=\frac{10087}{81}\\=124.53086~(approximate)$

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2 years ago

Which expression is equivalent to 5x + 10 y - 15

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2 years ago

Can anyone help me!

zhenek [66]

Answer:

$\large \text{$ \sf 2^{-64}$}$

Step-by-step explanation:

Given:

$\large \text{$ \sf (256)^{-4^{\frac{3}{2}}}$}$

This reads as: 256 to the power of "-4 to the power of 3/2".

Therefore, we need to deal with the "-4 to the power of 3/2" first.

Rewrite the 4 as 2²:

$\large \text{$ \sf \implies -(2^2)^{\frac{3}{2}}$}$

$\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\large \text{$ \sf \implies -2^{\left(2 \times \frac{3}{2}\right)}$}$

$\large \text{$ \sf \implies -2^{3}$}$

Therefore:

$\large \text{$ \sf \implies -2^3=(-2)(-2)(-2) = -8$}$

Replace "-4 to the power of 3/2" with -8 :

$\large \text{$ \sf \implies 256^{-8}$}$

Rewrite 256 as 2⁸ :

$\large \text{$ \sf \implies (2^8)^{-8}$}$

$\textsf{Again, apply exponent rule} \quad (a^b)^c=a^{bc}:$

$\large \text{$ \sf \implies 2^{(8 \times -8)}$}$

$\large \text{$ \sf \implies 2^{-64}$}$

In one complete calculation:

$\large\begin{aligned} \sf (256)^{-4^{\frac{3}{2}}} & = \sf (256)^{-(2^2)^{\frac{3}{2}}}\\& = \sf (256)^{-2^{\left(2 \times \frac{3}{2}\right)}}\\& =\sf (256)^{-2^{3}}\\& = \sf (256)^{-8}\\& \sf = (2^8)^{-8}\\& = \sf 2^{(8 \times -8)}\\& =\sf 2^{-64}\end{aligned}$