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3 years ago

To find the standard equation for a circle centered at the origin, we use the distance formula, since the radius measures?

Mathematics

2 answers:

Mrac [35]3 years ago

3 0

The distance from any point on the circle to the origin

____ [38]3 years ago

3 0

Answer: The answer is (a). the distance from any point in the circle to the origin.

Step-by-step explanation: We know that to find the standard equation of a circle centred at the origin, we always use the distance formula. We are to select the correct measurement for the radius from the options given.

In a circle, as shown in the attached figure, the radius 'r' is the distance from any point 'P' on the circle to the centre 'O' of the circle. The centre of the circle in standard form is usually the origin.

So, we can say that radius measures the distance from any point on the circle to the origin.

Thus, (a) is the correct option.

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3 years ago

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3 years ago

Which of the following is an extraneous solution of (45-3x)^1/2=x-9?

Iteru [2.4K]

Answer:

C. x =3

Step-by-step explanation:

Extraneous solution is that root of a transformed equation that doesn't satisfy the equation in it's original form because it was excluded from the domain of the original equation.

Let's solve the equation first

$\sqrt{45-3x} = x-9\\Taking\ square\ on\ both\ sides\\{(\sqrt{45-3x})}^2 = {(x-9)}^2\\45-3x = x^2-18x+81\\0 = x^2-18x+81-45+3x\\x^2-15x+36 = 0\\x^2-12x-3x+36 = 0\\x(x-12)-3(x-12) = 0\\(x-3)(x-12)\\x-3 = 0\\=> x =3\\x-12 = 0\\x = 12\\We\ will\ check\ the\ solutions\ one\ by\ one\\So,\\for\ x=3\\\sqrt{45-3(3)} = 3-9\\\sqrt{45-9} = -6\\\sqrt{36}= -6\\6\neq -6\\For x=12\\\sqrt{45-3(12)} = 12-9\\\sqrt{45-36} = 6\\\sqrt{36}= 6\\6=6$

Hence, we can conclude that x=3 is an extraneous solution of the equation ..

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4 years ago

What is the equation of a line that its parallel to -3x+4y=4 and passes through the point (4,0)?

anzhelika [568]

Answer:

y=3/4x-3

Step-by-step explanation:

Okay so to be parallel to an equation of another line, the line must have the same slope but a different y-intercept. First, you need to find the slope of the equation -3x+4y=4. The equation must be in y=mx+b format, so you first move -3x to the right by adding it on both sides, leaving you with 4y=3x+4. Next, you divide 4y by 4 in order to isolate y and do that on the other side. Now, you have y=3/4x+1.

To find the equation of the new line, you must put the point and slope into point slope form: y-y1=m(x-x1). In the point (4, 0), 4 would be x1 and 0 would be y1. so, the new equation is y-0=3/4 (x-4). now, distrubute the 3/4, leaving you with y=3/4x-3. This is correct since the slope remains the same in both equations with a different y-intercept. To check deeper, you could place point (4, 0) on a graph and then rise 3, run 4 until you get the line :) hope this helps

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3 years ago