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Fofino [41]
2 years ago
11

In ΔPQR, p = 75 inches, q = 73 inches and r=35 inches. Find the measure of ∠P to the nearest 10th of a degree.

Mathematics
2 answers:
love history [14]2 years ago
5 0

Answer:

cosp = 79.5

Step-by-step explanation:

soldi70 [24.7K]2 years ago
3 0

Answer:

79.5

Step-by-step explanation:

a^2 =b^2+ c^2-2bc cos A

cos p = 73^2+35^2-75^2/2(73)(35)

cos p= 9229/5110

cosp=79.5

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Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \  \theta  d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \   d \theta

A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  \dfrac{cos \ 2 \theta +1}{2}  \end {pmatrix} \ \ d \theta - \dfrac{25}{2}  \begin {bmatrix} \theta   \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}

A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix}  {cos \ 2 \theta +1}  \end {pmatrix} \ \    d \theta - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}}    \ \ - \dfrac{25}{2}  \begin {bmatrix}  \dfrac{2 \pi}{3} \end {bmatrix}

A =25  \begin {bmatrix}  \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3})  \end {bmatrix} - \dfrac{25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} +   \dfrac{\pi}{3}  \end {bmatrix}- \dfrac{ 25 \pi}{3}

A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
7 0
3 years ago
A pole that is 2.8 mm tall casts a shadow that is 1.37 m long. At the same time, a nearby building casts a shadow that is 36.25
ZanzabumX [31]

Answer:

this is the correct answer

5 0
2 years ago
Read 2 more answers
Find the side of to cube<br> whose volume is<br> 5.832cm3
Scorpion4ik [409]

Answer:

1.8cm

Step-by-step explanation:

Use the volume of cube formula.

\displaystyle V=s^3

\displaystyle5.832=s^3

\displaystyle\sqrt[3]{5.832} =s

\displaystyle s=1.8

3 0
3 years ago
Read 2 more answers
One hundred times the quantity six plus five
Andru [333]
100 times (6+5)
100 times 11
100 groups of 11=1100
Hope I helped:)

3 0
3 years ago
Read 2 more answers
I GIVE BRAINLIEST! LOOK AT THE IMAGE BELOW.
MissTica
Answer: 8

Area of a triangle=(1/2)bh
16=(1/2)4x
16=2x
x=8
3 0
3 years ago
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