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zaharov [31]
3 years ago
6

The ratio of the circumference of two circles is 3:2.What is the ratio of their area

Mathematics
1 answer:
Nezavi [6.7K]3 years ago
5 0

The ratio of their area is 9/4

<h2>Explanation:</h2>

We use ratios to compares values. In this exercise, we are comparing the circumference of two circles, which is:

3:2

and we want to know what is the ratio of their area. Recall that the circumference of a circle is given by:

C=2\pi r \\ \\ \\ Where: \\ \\ C:Circumference \\ \\ r:Radius \ of \ the \ circle

If we define:

C_{1}: \ Circumference \ of \ circle \ 1 \\ \\ C_{2}: \ Circumference \ of \ circle \ 2 \\ \\ r_{1}: \ Radius \ of \ circle \ 1 \\ \\ r_{2}: \ Radius \ of \ circle \ 2

Then, the ratio of the circumference of two circles is 3:2 is:

\frac{C_{1}}{C_{2}}=\frac{2\pi r_{1}}{2\pi r_{2}}=\frac{3}{2} \\ \\ \therefore \frac{r_{1}}{r_{2}}=\frac{3}{2}

The area of a circle is given by:

A=\pi r^2 \\ \\ A:Area \\ \\ r:Radius

So the ratio of their area can be found as:

\frac{A_{1}}{A_{2}}=\frac{\pi r_{1}^2}{\pi r_{2}^2} \\ \\ \\ A_{1}:Area \ of \ circle \ 1 \\ \\ A_{2}:Area \ of \ circle \ 2

So:

\frac{A_{1}}{A_{2}}=\frac{r_{1}^2}{r_{2}^2}=\left( \frac{r_{1}}{r_{2}} \right)^2 \\ \\ \frac{A_{1}}{A_{2}}=\left(\frac{3}{2}\right)^2 \\ \\ \boxed{\frac{A_{1}}{A_{2}}=\frac{9}{4}}

<h2>Learn more:</h2>

Unit rate: brainly.com/question/13771948#

#LearnWithBrainly

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find the parametric equations for the line of intersection of the two planes z = x + y and 5x - y + 2z = 2. Use your equations t
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Answer:

You didn't give the points in which you want the parametric equations be filled, but I have obtained the parametric equations, and they are:

x = (1/3 + t)

y = (-1/3 - 7t)

z = -6t

Step-by-step explanation:

If two planes intersect each other, the intersection will always be a line.

The vector equation for the line of intersection is given by

r = r_0 + tv

where r_0 is a point on the line and v is the vector result of the cross product of the normal vectors of the two planes.

The parametric equations for the line of intersection are given by

x = ax, y = by, and z = cz

where a, b and c are the coefficients from the vector equation

r = ai + bj + ck

To find the parametric equations for the line of intersection of the planes.

x + y - z = 0

5x - y + 2z = 2

We need to find the vector equation of the line of intersection. In order to get it, we’ll need to first find v, the cross product of the normal vectors of the given planes.

The normal vectors for the planes are:

For the plane x + y - z = 0, the normal vector is a⟨1, 1, -1⟩

For the plane 5x - y + 2z = 2, the normal vector is b⟨5, -1, 2⟩

The cross product of the normal vectors is

v = a × b =

|i j k|

|1 1 -1|

|5 -1 2|

= i(2 - 1) - j(2 + 5) + k(-1 - 5)

= i - 7j - 6k

v = ⟨1, -7, -6⟩

We also need a point on the line of intersection. To get it, we’ll use the equations of the given planes as a system of linear equations. If we set z = 0 in both equations, we get

x + y = 0

5x - y = 2

Adding these equations

5x + x + y - y = 2 + 0

6x = 2

x = 1/3

Substituting x = 1/3 back into

x + y = 0

y = -1/3

Putting these values together, the point on the line of intersection is

(1/3, -1/3, 0)

r_0= (1/3) i - (1/3) j + 0 k

r_0​​ = ⟨1/3, -1/3, 0⟩

Now we’ll plug v and r_0​​ into the vector equation.

r = r_0​​ + tv

r = (1/3)i - (1/3)j + 0k + t(i - 7j - 6k)

= (1/3 + t) i - (1/3 + 7t) j - 6t k

With the vector equation for the line of intersection in hand, we can find the parametric equations for the same line. Matching up r = ai + bj + ck with our vector equation,

r = (1/3 + t) i + (-1/3 - 7t) j + (-6t) k

a = (1/3 + t)

b = (-1/3 - 7t)

c = -6t

Therefore, the parametric equations for the line of intersection are

x = (1/3 + t)

y = (-1/3 - 7t)

z = -6t

3 0
3 years ago
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