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3 years ago

Find the radius of the two other circles

Mathematics

1 answer:

Lena [83]3 years ago

6 0

Answer:

We draw the circles on a coordinate system. We know that a circle is has points that are equidistant from its centerpoint. So the point where these circles are tangent is the midpoint between the centerpoints.

We can use the midpoint formula.

M = ((x1 + x2) / 2 , (y1 + y2) / 2)

M = ((-2 + 6) / 2 , (3 + 9) / 2)

M = (2, 6)

These circles will have the same radii.

If we take the distance between the midpoint and (-2,3), we will have the radius of the circle with this centerpoint.

If we take the distance between the midpoint and (6,9), we will have the radius of the circle with this centerpoint.

Step-by-step explanation:

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If anyone could help me with this last problem, I would greatly appreciate it.

il63 [147K]

Answer:

From the given information, the value of a is 3 and the measurement of ∠R is 25°

Step-by-step explanation:

For this problem, we have to find the value of a and the measurement of ∠R. We are given some information already in the problem.

This means that all of the angles and all of the sides of each triangle are going to be equal to each other.

So, knowing this, let;s find the measurement of ∠R first.

All triangles have a total measurement of 180°. We are already given two angle measurements. We are given that the m∠P is 90° because the small box in the triangle represents a right angle and right angles equal 90°. We are also given that the m∠Q is 65° because ∠Q is equal to ∠K so they have the same measurement. Now, let's set up our equation.

65 + 90 + m∠R = 180

Add 65 to 90.

155 + m∠R = 180

Subtract 155 from 180.

m∠R = 25°

So, the measurement of ∠R is 25°.

Now let's find the value of a.

KL is equal to PQ so we will set up an equation where they are equal to each other.

7a - 10 = 11

Add 10 to 11.

7a = 21

Divide 7 by 21.

a = 3

So, the value of a is 3.

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4 years ago

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3 years ago

Find the area of the surface. The part of the sphere x2 + y2 + z2 = a2 that lies within the cylinder x2 + y2 = ax and above the

sineoko [7]

Answer:

The area of the sphere in the cylinder and which locate above the xy plane is $\mathbf{ a^2 ( \pi -2)}$

Step-by-step explanation:

The surface area of the sphere is:

$\int \int \limits _ D \sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 ) } \ dA$

and the cylinder $x^2 + y^2 =ax$ can be written as:

$r^2 = arcos \theta$

$r = a cos \theta$

where;

D = domain of integration which spans between $\{(r, \theta)| - \dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}, 0 \leq r \leq acos \theta\}$

and;

the part of the sphere:

$x^2 + y^2 + z^2 = a^2$

making z the subject of the formula, then :

$z = \sqrt{a^2 - (x^2 +y^2)}$

Thus,

$\dfrac{\partial z}{\partial x} = \dfrac{-2x}{2 \sqrt{a^2 - (x^2+y^2)}}$

$\dfrac{\partial z}{\partial x} = \dfrac{-x}{ \sqrt{a^2 - (x^2+y^2)}}$

Similarly;

$\dfrac{\partial z}{\partial y} = \dfrac{-2y}{2 \sqrt{a^2 - (x^2+y^2)}}$

$\dfrac{\partial z}{\partial y} = \dfrac{-y}{ \sqrt{a^2 - (x^2+y^2)}}$

So;

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\begin {pmatrix} \dfrac{-x}{\sqrt{a^2 -(x^2+y^2)}} \end {pmatrix}^2 + \begin {pmatrix} \dfrac{-y}{\sqrt{a^2 - (x^2+y^2)}} \end {pmatrix}^2+1}$ $\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{x^2+y^2}{a^2 -(x^2+y^2)}+1}$

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{x^2+y^2+a^2 -(x^2+y^2)}{a^2 -(x^2+y^2)}}$

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = \sqrt{\dfrac{a^2}{a^2 -(x^2+y^2)}}$

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = {\dfrac{a}{\sqrt{a^2 -(x^2+y^2)}}$

From cylindrical coordinates; we have:

$\sqrt{(\dfrac{\partial z}{\partial x})^2 + ( \dfrac{\partial z}{\partial y}^2 + 1 )} = {\dfrac{a}{\sqrt{a^2 -r^2}}$

dA = rdrdθ

By applying the symmetry in the x-axis, the area of the surface will be:

$A = \int \int _D \sqrt{ (\dfrac{\partial z}{\partial x})^2+ (\dfrac{\partial z}{\partial y})^2+1} \ dA$

$A = \int^{\dfrac{\pi}{2}}_{-\dfrac{\pi}{2}} \int ^{a cos \theta}_{0} \dfrac{a}{\sqrt{a^2 -r^2 }} \ rdrd \theta$

$A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 -r^2} \end {bmatrix}^{a cos \theta}_0 \ d \theta$

$A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \sqrt{a^2 - a^2cos^2 \theta} + a \sqrt{a^2 -0}} \end {bmatrix} d \theta$ $A = 2\int^{\dfrac{\pi}{2}}_{0} \begin {bmatrix} -a \ sin \theta +a^2 } \end {bmatrix} d \theta$