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Rainbow [258]
3 years ago
6

What is the area of a trapezoid with one base being 10 units, another base being 14 units, and a height of 6 units?

Mathematics
1 answer:
kolbaska11 [484]3 years ago
4 0

Answer:

A=72

Step-by-step explanation:

A=a+b

2h=10+14

2·6=72

A=72

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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
Please answer this correctly
borishaifa [10]

Answer:

47

Step-by-step explanation:

The probability for tails to land is 1/2 when you flip 1 coin.

When you flip 94 times, 1/2 × 94

= 47

6 0
3 years ago
Solve: 1/4x - 5 = 3/4x - 12
ivanzaharov [21]
The answer to the math problem it X=14 hope this helps
7 0
3 years ago
Read 2 more answers
What is the simplified form of each expression? 4(20 + 12) ÷ (4 - 3)
kodGreya [7K]

Hi!

<h3>Let's solve this using PEMDAS.</h3><h3>Parentheses - Addition</h3>

4 * 32 ÷ (4 - 3)

<h3>Parentheses - Subtraction</h3>

4 * 32 ÷ 1

<h3>Multiplication</h3>

128 ÷ 1

<h3>Division</h3>

<u>128</u>

<h2>The answer is 128.</h2>

Hope this helps! :)

-Peredhel

7 0
3 years ago
What is the total balance for a loan of $400 compounded annually that has a rate of 11% for 4 years
AURORKA [14]

Answer: 579.39

Step-by-step explanation:

P×R×T

400×11%×31/365 to get interest rate per month= 3.737

Total interest- 3.737×12×4= 179.38

Total interest plus principal =Total balance = 579.39

4 0
3 years ago
Read 2 more answers
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