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kicyunya [14]
2 years ago
15

The figure of Circle A shown below has diameter PR which intersects QS at point B and the measurements shown. Calculate the foll

owing measures:
m

Complete all calculations and work on a separate piece of paper, including any figures that you use. Once you have found all of the measures, upload the solution and your work here.


Please answer ASAP!!! 35 POINTS!!!

Mathematics
1 answer:
nata0808 [166]2 years ago
6 0

The figure is an illustration of the relationship between the angles in a circle and arc

<h3>The measure of angle PSQ</h3>

From the figure, the arc PQ is subtended by the angle PAQ.

This means that:

PQ = ∠PAQ

Given that ∠PAQ = 130, it means that:

Arc PQ = 130

The measure of PSQ is then calculated using:

∠PSQ = 0.5 * Arc PQ ----- inscribed angle is half a subtended angle.

This gives

∠PSQ = 0.5 * 130

∠PSQ = 65

Hence, the measure of ∠PSQ is 65 degrees

<h3>The measure of arc QR</h3>

A semicircle measures 180 degrees.

This means that:

QR + PQ = 180

So, we have:

QR = 180 - PQ

Substitute 130 for PQ

QR = 180 - 130

QR = 50

Hence, the measure of QR is 50 degrees

<h3>The measure of arc RS</h3>

The measure of arc RS is then calculated using:

∠RPS = 0.5 * Arc RS ----- inscribed angle is half a subtended angle.

Where ∠RPS = 35

So, we have:

35 = 0.5 * Arc RS

Multiply both sides by 2

Arc RS = 70

Hence, the measure of RS is 70 degrees

<h3>The measure of angle AQS</h3>

In (a), we have:

∠PSQ = 65

This means that:

∠PSQ = ∠PSB = 65

So, we have:

∠PSB = 65

Next, calculate SBP using:

∠SBP + ∠BPS + ∠PSB = 180 ---- sum of angles in a triangle.

So, we have:

∠SBP + 35 + 65 = 180

∠SBP + 100 = 180

This gives

∠SBP = 80

The measure of AQS is then calculated using:

AQS = AQB = 180 - (180 - SBP) - (180 - PAQ)

This gives

AQS = 180 - (180 - 80) - (180 - 130)

Evaluate

AQS = 30

Hence, the measure of AQS is 30 degrees

<h3>The measure of arc PS</h3>

A semicircle measures 180 degrees.

This means that:

PS + RS = 180

This gives

PS = 180 - RS

Where RS = 70

So, we have:

PS = 180 - 70

Evaluate

PS = 110

Hence, the measure of arc PS is 110 degrees

Read more about circles and arcs at:

brainly.com/question/15096899

#SPJ1

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In this problem you will use undetermined coefficients to solve the nonhomogeneous equation y′′+4y′+3y=8te^−t+6e^−t−(9t+6)
Luden [163]

We're given the ODE,

<em>y''</em> + 4<em>y'</em> + 3<em>y</em> = 8<em>t</em> exp(-<em>t </em>) + 6 exp(-<em>t</em> ) - (9<em>t</em> + 6)

(where I denote exp(<em>x</em>) = <em>eˣ </em>)

First determine the characteristic solution:

<em>y''</em> + 4<em>y'</em> + 3<em>y</em> = 0

has characteristic equation

<em>r</em> ² + 4<em>r</em> + 3 = (<em>r</em> + 1) (<em>r</em> + 3) = 0

with roots at <em>r</em> = -1 and <em>r</em> = -3, so the characteristic solution is

<em>y</em> = <em>C</em>₁ exp(-<em>t</em> ) + <em>C</em>₂ exp(-3<em>t</em> )

For the non-homogeneous equation, assume two ansatz solutions

<em>y</em>₁ = (<em>at</em> ² + <em>bt</em> + <em>c</em>) exp(-<em>t </em>)

and

<em>y</em>₂ = <em>at</em> + <em>b</em>

<em />

• <em>y''</em> + 4<em>y'</em> + 3<em>y</em> = 8<em>t</em> exp(-<em>t </em>) + 6 exp(-<em>t</em> ) … … … [1]

Compute the derivatives of <em>y</em>₁ :

<em>y</em>₁ = (<em>at</em> ² + <em>bt</em> + <em>c</em>) exp(-<em>t </em>)

<em>y</em>₁' = (2<em>at</em> + <em>b</em>) exp(-<em>t </em>) - (<em>at</em> ² + <em>bt</em> + <em>c</em>) exp(-<em>t </em>)

… = (-<em>at</em> ² + (2<em>a</em> - <em>b</em>) <em>t</em> + <em>b</em> - <em>c</em>) exp(-<em>t </em>)

<em>y</em>₁'' = (-2<em>at</em> + 2<em>a</em> - <em>b</em>) exp(-<em>t </em>) - (-<em>at</em> ² + (2<em>a</em> - <em>b</em>) <em>t</em> + <em>b</em> - <em>c</em>) exp(-<em>t </em>)

… = (<em>at</em> ² + (<em>b</em> - 4<em>a</em>) <em>t</em> + 2<em>a</em> - 2<em>b</em> + <em>c</em>) exp(-<em>t</em> )

Substitute them into the ODE [1] to get

→   [(<em>at</em> ² + (<em>b</em> - 4<em>a</em>) <em>t</em> + 2<em>a</em> - 2<em>b</em> + <em>c</em>) + 4 (-<em>at</em> ² + (2<em>a</em> - <em>b</em>) <em>t</em> + <em>b</em> - <em>c</em>) + 3 (<em>at</em> ² + <em>bt</em> + <em>c</em>)] exp(-<em>t</em> ) = 8<em>t</em> exp(-<em>t </em>) + 6 exp(-<em>t</em> )

(<em>at</em> ² + (<em>b</em> - 4<em>a</em>) <em>t</em> + 2<em>a</em> - 2<em>b</em> + <em>c</em>) + 4 (-<em>at</em> ² + (2<em>a</em> - <em>b</em>) <em>t</em> + <em>b</em> - <em>c</em>) + 3 (<em>at</em> ² + <em>bt</em> + <em>c</em>) = 8<em>t</em> + 6

4<em>at</em> + 2<em>a</em> + 2<em>b</em> = 8<em>t</em> + 6

→   4<em>a</em> = 8   and   2<em>a</em> + 2<em>b</em> = 6

→   <em>a</em> = 2   and   <em>b</em> = 1

→   <em>y</em>₁ = (2<em>t</em> ² + <em>t </em>) exp(-<em>t </em>)

(Note that we don't find out anything about <em>c</em>, but that's okay since it would have gotten absorbed into the first characteristic solution exp(-<em>t</em> ) anyway.)

• <em>y''</em> + 4<em>y'</em> + 3<em>y</em> = -(9<em>t</em> + 6) … … … [2]

Compute the derivatives of <em>y</em>₂ :

<em>y</em>₂ = <em>at</em> + <em>b</em>

<em>y</em>₂' = <em>a</em>

<em>y</em>₂'' = 0

Substitute these into [2] :

4<em>a</em> + 3 (<em>at</em> + <em>b</em>) = -9<em>t</em> - 6

3<em>at</em> + 4<em>a</em> + 3<em>b</em> = -9<em>t</em> - 6

→   3<em>a</em> = -9   and   4<em>a</em> + 3<em>b</em> = -6

→   <em>a</em> = -3   and   <em>b</em> = 2

→   <em>y</em>₂ = -3<em>t</em> + 2

Then the general solution to the original ODE is

<em>y(t)</em> = <em>C</em>₁ exp(-<em>t</em> ) + <em>C</em>₂ exp(-3<em>t</em> ) + (2<em>t</em> ² + <em>t </em>) exp(-<em>t </em>) - 3<em>t</em> + 2

Use the initial conditions <em>y</em> (0) = 2 and <em>y'</em> (0) = 2 to solve for <em>C</em>₁ and <em>C</em>₂ :

<em>y</em> (0) = <em>C</em>₁ + <em>C</em>₂ + 2 = 2

→   <em>C</em>₁ + <em>C</em>₂ = 0 … … … [3]

<em>y'(t)</em> = -<em>C</em>₁ exp(-<em>t</em> ) - 3<em>C</em>₂ exp(-3<em>t</em> ) + (-2<em>t</em> ² + 3<em>t</em> + 1) exp(-<em>t </em>) - 3

<em>y'</em> (0) = -<em>C</em>₁ - 3<em>C</em>₂ + 1 - 3 = 2

→   <em>C</em>₁ + 3<em>C</em>₂ = -4 … … … [4]

Solve equations [3] and [4] to get <em>C</em>₁ = 2 and <em>C</em>₂ = -2. Then the particular solution to the initial value problem is

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