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svp [43]
3 years ago
13

25/8(2/3+8/3(5/4-6/5)) Worth 10 points pls help

Mathematics
1 answer:
melamori03 [73]3 years ago
8 0
Here is your answer. try to solve it using Bodmas rule

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A) FIGURE 1

I HOPE YOU WILL LIKE IT?:-)

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Which of the following could be the side measures of a right triangle?
salantis [7]

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D

Step-by-step explanation:

16+9=25 so it's valid

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Find the length and area of a rectangular rug if the width of the rug is 12 feet and the diagonal measures 20 feet.
Natalka [10]

Answer:

So use pythogoras theorem

a^2=c^2-b^2

a^2 = 20^2-12^2

a^2= 256

a= square root of 256

length is 16 ft

area = l x w

= 16 x 12

Area is 192 square feet

Step-by-step explanation:

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Find the total area of the irregular polygon shown? * m 11 m 36 sqm 84 sq.m 88 sqm​
VARVARA [1.3K]

Answer:

100

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Split it into 3 different rectangles, find the area of each, and add them together.

Rectangle 1: length=11   width=4    11×4=44

Rectangle 2: length=11-9=2   width=6    2×6=12

Rectangle 3: length=11   width=4    11×4=44

44+12+44=100

3 0
3 years ago
Simplify csc θ(1−cos^2 θ)/sinθ cos θ to a single trigonometric function
topjm [15]

Answer:

<u><em></em></u>\frac{cosec\alpha(1-cosx^{2} \alpha ) }{sin\alpha cos\alpha  } = sec\alpha<u><em></em></u>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the trigonometric function

            \frac{cosec\alpha(1-cosx^{2} \alpha ) }{sin\alpha cos\alpha  }

  we know that    sin²∝ + cos²∝ = 1

                      ⇒    sin²∝ = 1- cos²∝

Now we have to simplify the given trigonometric function

                =       \frac{cosec\alpha(sin^{2} \alpha  ) }{sin\alpha cos\alpha  }

               =       \frac{cosec\alpha(sin \alpha  ) }{ cos\alpha  }

<u><em>Step(ii)</em></u>

         we know that  cosec∝ = 1/ sin∝

              = \frac{\frac{1}{sin\alpha } (sin \alpha  ) }{ cos\alpha  }

    After cancellation sine function, we get

          =  \frac{1 }{ cos\alpha  } = sec\alpha

<u><em>Final answer:-</em></u>

<u><em></em></u>\frac{cosec\alpha(1-cosx^{2} \alpha ) }{sin\alpha cos\alpha  } = sec\alpha<u><em></em></u>

           

     

 

               

6 0
3 years ago
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