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sdas [7]
3 years ago
6

In circle L, points E and F lie on the circle such that E, F and L are not collinear. If LE, LF and EF are drawn then which of t

he following must be true about LEF?
(1) It is a right triangle.
(2) it is and isosceles triangle.
(3) it is an equilateral triangle.
(4) it is an obtuse triangle.​
Mathematics
1 answer:
frozen [14]3 years ago
8 0

Answer:

<u>option (2) it is and isosceles triangle. </u>

Step-by-step explanation:

In circle L, points E and F lie on the circle such that E, F and L are not collinear

If LE, LF and EF are drawn

So, LE = EF = the radius of the circle L

So, LEF is an isosceles triangle.

The answer is option (2) it is and isosceles triangle.

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Answer:

17.8

Step-by-step explanation:

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9.5x = 169

x = 169/9.5

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y is directly proportional to x^3. it is known that =5 for a particular value of x. find the value of y when this value of y whe
alina1380 [7]

Answer:

The value of y when the value of x is multiplied by \frac{1}{2} is \frac{5}{8}.

Step-by-step explanation:

According to the statement, we have the following relationship:

y = k\cdot x^{3} (1)

Where:

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y - Dependent variable.

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We can eliminate the proportionality constant by constructing the following relationship:

\frac{y_{2}}{y_{1}} = \left(\frac{x_{2}}{x_{1}} \right)^{3}

If we know that y_{1} = y, y_{2} = 5, x_{2} = x_{o} and x_{1} = \frac{1}{2}\cdot x_{o}, then the value of y when the value of x is multiplied by \frac{1}{2} is:

\frac{5}{y} = \left(\frac{x_{o}}{\frac{1}{2}\cdot x_{o} } \right)^{3}

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The value of y when the value of x is multiplied by \frac{1}{2} is \frac{5}{8}.

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