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dmitriy555 [2]
3 years ago
7

Can u guys answer my question real quick pls

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
4 0

Answer:

  • 11 = 103
  • 12 = 1.732

Step-by-step explanation:

\frac{1}{216^{\frac{-2}{3}}}+\frac{1}{256^{-\frac{3}{4}}}+\frac{1}{243^{-\frac{1}{5}}}\\\\Let \: start  \:with ; \: \frac{1}{216^{\frac{-2}{3}}}\\\\\mathrm{Apply\:the\:fraction\:rule}:\\\quad \frac{-a}{b}=-\frac{a}{b}\\=\frac{1}{216^{-\frac{2}{3}}}\\216^{-\frac{2}{3}}\\\\\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}\\=\frac{1}{216^{\frac{2}{3}}}\\\\216^{\frac{2}{3}}\\=\left(6^3\right)^{\frac{2}{3}}\\\left(6^3\right)^{\frac{2}{3}}=6^{3\times \frac{2}{3}}=6^2\\=36=\frac{1}{\frac{1}{36}}\\

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{1}{\frac{b}{c}}=\frac{c}{b}=\frac{36}{1}=36\\\frac{1}{256^{-\frac{3}{4}}}\\\\256^{-\frac{3}{4}}=\frac{1}{64}\\\\=\frac{1}{\frac{1}{64}} = 64/1 = 6\\\\\frac{1}{243^{-\frac{1}{5}}}\\\\243^{-\frac{1}{5}}=\frac{1}{3}\\\\=\frac{1}{\frac{1}{3}} = 3/1 = 3\\\\\\=36+64+3\\=103

\frac{3+\sqrt{6}}{5\sqrt{3}-2\sqrt{12}-\sqrt{32}+\sqrt{50}}\\\\2\sqrt{12} = =4\sqrt{3}\\\\\sqrt{32}=4\sqrt{2}\\\\\sqrt{50}=5\sqrt{2}\\\\=\frac{3+\sqrt{6}}{5\sqrt{3}-4\sqrt{3}-4\sqrt{2}+5\sqrt{2}}\\\\5\sqrt{3}-4\sqrt{3}-4\sqrt{2}+5\sqrt{2}=\sqrt{3}+\sqrt{2}\\\\\mathrm{Add\:similar\:elements:}\:-4\sqrt{2}+5\sqrt{2}=\sqrt{2}\\\\=5\sqrt{3}-4\sqrt{3}+\sqrt{2}\\\\\mathrm{Add\:similar\:elements:}\:5\sqrt{3}-4\sqrt{3}=\sqrt{3}\\\\

\mathrm{Add\:similar\:elements:}\:5\sqrt{3}-4\sqrt{3}=\sqrt{3}\\\\=\sqrt{3}+\sqrt{2}\\\=\frac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}\\\mathrm{Multiply\:by\:the\:conjugate}\:\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}\\\\=\frac{\left(3+\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)}\\{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\\\\\left(3+\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)=\sqrt{3}\\\\\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=1\\=\frac{\sqrt{3}}{1}\\\\=\sqrt{3}

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The extensions of the legs AB and CD of a trapezoid ABCD intersect at point E. Find the lengths of the sides of △AED if AB=5cm,
Lynna [10]

Answer:

  AE = 15 cm; ED = 18 cm; AD = 15 cm (given)

Step-by-step explanation:

ΔBEC ~ ΔAED so ...

   AD/BC = AE/BE = (BE+AB)/BE = 1 + AB/BE

Substituting given numbers (lengths in centimeters), we have ...

  15/10 = 1 + 5/BE

  1/2 = 5/BE

  BE = 10

Similarly, ...

  1/2 = 6/CE

  CE = 12

Then the unknown sides are ...

  AE = AB + BE = 5 + 10 = 15 . . . cm

  ED = CE + CD = 12 + 6 = 18 . . . cm

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3 years ago
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Answer:

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Step-by-step explanation:

Use the compound interest formula and solve for t. Logarithms are involved.

  A = P(1 +r/n)^(nt)

amount when P is invested for t years at annual rate r compounded n times per year.

Using the given values, we have ...

  13060 = 8800(1 +0.028/365)^(365t)

  13060/8800 = (1 +0.028/365)^(365t) . . . . divide by P=8800

Now we take logarithms to make this a linear equation.

  log(13060/8800) = (365t)log(1 +0.028/365)

Dividing by the coefficient of t gives us ...

  t = log(13060/8800)/(365·log(1 +0.028/365)) ≈ 0.171461/0.0121598

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It would take about 14.1 years for the value to reach $13,060.

8 0
3 years ago
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Answer:

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Step-by-step explanation:

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