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Lana71 [14]
2 years ago
13

NEED HELP ASAP (please and thank you)

Mathematics
2 answers:
vlada-n [284]2 years ago
8 0

Answer:800

i put it wrong at first but its 800

solmaris [256]2 years ago
7 0

Answer:

800

Step-by-step explanation:

i dont know how to explain. i just did the math. hope this helps.

it wouldnt be 600 because thats 1/2 of 1200 not 2/3.  

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Find the sum of -6/ab + a^2/b^2
Ymorist [56]

Answer:

\frac{1}{ab^{2} } (a³ - 6b)

Step-by-step explanation:

\frac{-6}{ab} + \frac{a^{2} }{b^{2} }

= \frac{-6b + a^{3} }{ab^{2} }

= \frac{1}{ab^{2} } (a³ - 6b)

4 0
3 years ago
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
Which set of shapes could be used to form a net for a square pyramid?
bulgar [2K]

Answer:

A.1 square and 4 triangles

Step-by-step explanation:

A pyramid has sides that are triangular faces and a base. In a square pyramid, the base is a square.

The net therefore has 1 Square and 4 Triangles.

A net is given in the attached diagram:

8 0
3 years ago
What is 11 feet and 8 inches plus 2 feet and 9 inches
Agata [3.3K]

Answer:

14 feet and 5 inches

Step-by-step explanation:

11 feet + 2 feet= 13 feet

9 inches + 8 inches = 17 inches

17 inches = 1 foot and 5 inches

13 + 1 = 14 feet

5 inches + 0 inches = 5 inches

7 0
3 years ago
What’s the quotient of 20 and 2
olasank [31]

the answer is 10 hoped this helped have a great day


5 0
3 years ago
Read 2 more answers
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