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ladessa [460]
2 years ago
13

Which one of these points lies on the given circle

Mathematics
1 answer:
Kruka [31]2 years ago
3 0

Step-by-step explanation:

every point on the circle has the same distance from the center : r (radius).

we can see that radius right there on the x-axis : the distance of (3, 0) to (0, 0) is simply 3. that is our radius.

the distance formula is applied Pythagoras

distance² = (difference in x coordinates)² +

+ (difference in y coordinates)²

distance = sqrt( ... )

so,

(-2.5, - 2.5)

(-2.5 - 0)² + (-2.5 - 0)² = 6.25 + 6.25 = 12.5

that is not 3² = 9, so the point is NOT on the circle.

(1, 3)

(1 - 0)² + (3 - 0)² = 1 + 9 = 10

that is not 3² = 9, so the point is NOT on the circle.

(2, -sqrt(5))

(2 - 0)² + (-sqrt(5) - 0)² = 4 + 5 = 9

that IS 3² = 9, so, the point IS on the circle.

(-sqrt(3), -sqrt(7))

(-sqrt(3) - 0)² + (-sqrt(7) - 0)² = 3 + 7 = 10

that is not 3² = 9, so the point is NOT on the circle.

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In order to solve for parallel, perpendicular, or neither, you have to look at the slope.

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If it's the reciprocal (Where you flip the number and add change the signs. For example, the reciprocal of 1/2 is -2)
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So for the first equation, your slope is:
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For the second equation, your slope is -3/2 since y=-3/2x+5 is already in y=mx+b form and m is the slope.

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Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=5x-%20%5Cdfrac%7B%20%5Csqrt%7B%209%20%7B%20x%20%7D%5E%7B%202%20%7D%20-6x%2B1%20%5Cphantom%7B%5
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Answer:

=5x+1          or          =5x-1

Step-by-step explanation:

One is given the following equation:

5x-\frac{\sqrt{9x^2-6x+1}}{1-3x}

The problem asks one to simplify the expression, the first step in solving this equation is to factor the equation.  Rewrite the numerator and denominator of the fraction as the product of two expressions. Remember the factoring patterns:

(a-b)^2=a^2-2ab+b^2

=5x-\frac{\sqrt{9x^2-6x+1}}{1-3x}

=5x-\frac{\sqrt{(3x-1)^2}}{-(3x-1)}

Now simplify the numerator. Remember, taking the square root of a squared value is the same as taking the absolute value of the expression,

=5x-\frac{\sqrt{(3x-1)^2}}{1-3x}

=5x-\frac{|3x-1|}{-(3x-1)}

Rewrite the expression without the absolute value sign in the numerator. Remember the general rule for removing the absolute value sign:

|a-b|\\=a -b           or           (-a-b) = b-a

=5x-\frac{|3x-1|}{-(3x-1)}

=5x-\frac{3x-1}{-(3x-1)}          or          =5x-\frac{-(3x-1)}{-(3x-1)}

Simplify both expressions, reduce by canceling out common terms in both the numerator and the denominator,

=5x-\frac{3x-1}{-(3x-1)}          or          =5x-\frac{-(3x-1)}{-(3x-1)}

=5x-(-1)          or          =5x-(1)

Simplify further by rewriting the expression without the parenthesis, remember to distribute the sign outside the parenthesis by the terms inside of the parenthesis; note that negative times negative equals positive.

=5x-(-1)          or          =5x-(1)

=5x+1          or          =5x-1

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