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3 years ago

9

−6x − 2y + 2z = −8 3x − 2y − 4z = 8 6x − 2y − 6z = −18

1 answer:

Troyanec [42]3 years ago

5 0

What is the question here

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The equation of a circle whose center is at ( 4,0 ) and radius is length 2 ( square root ) 3

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The equation of a circle:
$(x-h)^2+(y-k)^2=r^2$
(h,k) - the coordinates of the center
r - the radius

The center is (4,0), the length of the radius is 2√3.
$(x-4)^2+(y-0)^2=(2\sqrt{3})^2 \\
\boxed{(x-4)^2+y^2=12}$

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3 years ago

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Tasya [4]

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Step-by-step explanation:

$\sqrt[4]{(16a^4)^5}=(16a^4)^{\frac{5}{4}}$ [Since, $\sqrt[n]{x}=x^{\frac{1}{n}}$ ]

$=(16)^{\frac{5}{4}}(a^4)^{\frac{5}{4}}$

$=(2^4)^{\frac{5}{4}}(a^4)^{\frac{5}{4} }$

$=(2^{4\times \frac{5}{4}})(a^{4\times \frac{5}{4}})$

$=2^5a^5$

$\sqrt[5]{(b^{\frac{1}{4}})^6}=\sqrt[5]{b^{\frac{6}{4}}}$

$=\sqrt[5]{b^{\frac{3}{2}}}$

$=(b^{\frac{3}{2}})^{\frac{1}{5} }$

$=b^{\frac{3}{10}}$

$\sqrt[3]{\frac{c^{15}}{c^9}}=\sqrt[3]{c^{15-9}}$

$=(c^6)^{\frac{1}{3}}$

$=c^{6\times \frac{1}{3} }$

= c²

$\sqrt[4]{16d^6\times 16d^{-5}}=\sqrt[4]{(2)^4(d^{6-5})(2)^4}$

$=\sqrt[4]{2^{(4+4)}d}$

$=\sqrt[4]{(2^8)d}$

$=2^{\frac{8}{4}}d^{\frac{1}{4}}$

$=2^2d^{\frac{1}{4}}$

$=4d^{\frac{1}{4}}$

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3 years ago