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Basile [38]
3 years ago
12

Please help me no one is helping me and I need help :(

Mathematics
1 answer:
melamori03 [73]3 years ago
5 0

Answer:

JK is longer

Step-by-step explanation:

You can kinda tell just by looking at it. Chord JK is closer to the center of the circle so it's longer.

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Using substitution, which value of x makes this equation true?
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A. Find y in terms of x if
givi [52]

Answer:

A.y(x) =\sqrt{ 2(\frac{x^5}{5} -\frac{6^5}{5})}

B.Therefore the solution is defined on the interval:

6≤ x ≤ +∞

Step-by-step explanation:

Given differential equation is

\frac{dy}{dx} = x^4y^{-1}

\Rightarrow \frac{dy}{dx} =\frac{ x^4}{y}

\Rightarrow y dy = x^4dx

Integrating both sides:

\Rightarrow\int y dy = \int x^4dx

\Rightarrow \frac{y^2}{2} =\frac{x^5}{5} +C........(1)

Given y(0)= 6

it means y=0 when x=6

Putting x=6 and y=0 in the equation (1)

\therefore \frac{0^2}{2} =\frac{6^5}{5} +C

\Rightarrow C=-\frac{6^5}{5}

Equation (1) become:

\Rightarrow \frac{y^2}{2} =(\frac{x^5}{5} -\frac{6^5}{5})

\Rightarrow y} =\sqrt{ 2(\frac{x^5}{5} -\frac{6^5}{5})}

Therefore y(x) =\sqrt{ 2(\frac{x^5}{5} -\frac{6^5}{5})}

(B)

The quantity under root must greater than or equal to zero.

Therefore,

2(\frac{x^5}{5} -\frac{6^5}{5})} \geq0

\Rightarrow \frac{x^5}{5} -\frac{6^5}{5}} \geq0

\Rightarrow \frac{x^5}{5} \geq\frac{6^5}{5}}

\Rightarrow x^5\geq {6^5

\Rightarrow \sqrt[5]{x^5} \geq \sqrt[5]{6^5}

\Rightarrow x\geq {6

When x≥0 then the value of 2(\frac{x^5}{5} -\frac{6^5}{5})} is real other wise the value of 2(\frac{x^5}{5} -\frac{6^5}{5})} is imaginary.

Therefore the solution is defined on the interval:

6≤ x ≤ +∞

8 0
4 years ago
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