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3 years ago

Quiz

Mathematics

2 answers:

Tasya [4]3 years ago

7 0

Answer:

1603

Step-by-step explanation:

3^3*4^3-5^3

__________

3*3*3=9*3=27

4*4*4=16*4=64

5*5*5=25*5=125

(27*64)-125

1728-125

1603

Svetlanka [38]3 years ago

5 0

Answer:

Exponents =

given 3³ x 4³ - 5³

in formula :

$a^c × b^c = (a + b)^c$

Solution :

$3^3 × 4^3 - 5^3 = (3 \cdot 4)^3 - 5^3$

$3^3 × 4^3 - 5^3 = (12)^3 - 5^3$

$3^3 × 4^3 - 5^3 = 1728- 125$

$3^3 × 4^3 - 5^3 = 1603$

$\longmapsto \boxed{\bold {1603} \textsf{ In This Answer}}$

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3 years ago

Help with this integral<br><br><img src="https://tex.z-dn.net/?f=%20%5Cint%5Climits%20%5Cfrac%7Bdx%7D%7Bx%5E%7B2%7D-4x-13%7D" id

charle [14.2K]

$x^2-4x-13=(x-2)^2-17$

$x-2=\sqrt{17}\sec y$
$\mathrm dx=\sqrt{17}\sec y\tan y\,\mathrm dy$

$\displaystyle\int\frac{\mathrm dx}{x^2-4x-13}=\int\frac{\sqrt{17}\sec y\tan y}{(\sqrt{17}\sec y)^2-17}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\sec y\tan y}{\sec^2y-1}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\sec y\tan y}{\tan^2y}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\sec y}{\tan y}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\frac{\frac1{\cos y}}{\frac{\sin y}{\cos y}}\,\mathrm dy$
$=\displaystyle\frac1{\sqrt{17}}\int\csc y\,\mathrm dy$
$=-\dfrac1{\sqrt{17}}\ln|\csc y+\cot y|+C$

$\sec y=\dfrac{x-2}{\sqrt{17}}\iff y=\sec^{-1}\dfrac{x-2}{\sqrt{17}}$
$\implies\csc y=\dfrac{x-2}{\sqrt{(x-2)^2-17}}=\dfrac{x-2}{\sqrt{x^2-4x-13}}$
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$\displaystyle\int\frac{\mathrm dx}{x^2-4x-13}=-\dfrac1{\sqrt{17}}\ln\left|\frac{x-2+\sqrt{17}}{\sqrt{x^2-4x-13}}\right|+C$

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4 years ago