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jekas [21]
3 years ago
7

What is the approximate area of the shaded region? the shaded area is 28  A.168.56 square inches  B.696.08 square inches  C.740.

04 square inches  D.1677.76 square inches
Mathematics
1 answer:
svp [43]3 years ago
5 0
The answer is B its quite easy if you look at it
 
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3/8 x 5 as a mixed number​
leva [86]

Answer:

1 7/8

Step-by-step explanation:

3/8 * 5/1 = 15/8

<-- First multiply 3/8 by 5

----------------------------------------------

15/8 = 1 with a remainder of 7

<-- Then convert 15/8 to a mixed number by dividing 15 by 8

15/8 = 1 7/8

8 0
2 years ago
A simple one-celled organism divides into two identical
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24 single felled organisms there are to 30 minute periods in each hour which produces 2 organisms that is 12 periods and 12 multiplied by 2 is 24
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In this rotation WXY is mapped to CBA. if QON = 55, then XOB measures _____.
CaHeK987 [17]
The correct answer would be 1

8 0
3 years ago
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Solve - 1.2x + 2.5 = 0.8x + 6.5
lesantik [10]

Answer: x is -2

Steps:

- 1.2x + 2.5 =  0.8x + 6.5

- 1.2x - 0.8x = 6.5 - 2.5

- 2x = 4

x =  4/-2

x = - 2

Brainliest pls if I am correct!

6 0
2 years ago
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Find all the solutions of the equation in the interval [0,2pi). 4sin^(2)x=5-4cosx
gulaghasi [49]
                       4sin²(x) = 5 - 4cos(x)
        4{¹/₂[1 - cos(2x)]} = 5 - 4cos(x)
   4{¹/₂[1] - ¹/₂[cos(2x)]} = 5 - 4cos(x)
         4[¹/₂ - ¹/₂cos(2x)] = 5 - 4cos(x)
     4[¹/₂] - 4[¹/₂cos(2x)] = 5 - 4cos(x)
                2 - 2cos(2x) = 5 - 4cos(x)
              - 2                   - 2
                    -2cos(2x) = 3 - 4cos(x)
            -2[2cos²(x) - 1] = 3 - 4cos(x)
               -4cos²(x) + 2 = 3 - 4cos(x)
                               - 2  - 2
                     -4cos²(x) = 1 - 4cos(x)
-4cos²(x) + 4cos(x) - 1 = 0
 4cos²(x) - 4cos(x) + 1 = 0
               [2cos(x) - 1]² = 0
                  2cos(x) - 1 = 0
                              + 1 + 1
                       2cos(x) = 1
                            2         2
                         cos(x) = ¹/₂
                cos⁻¹[cos(x)] = cos⁻¹(¹/₂)
                                 x = 60, 300
                                 x = π/3, 5π/3

[0, 2π) = 0 ≤ x < 2π
[0, 2π) = 0 ≤ π/3 ≤ 2π or 0 ≤ 5pi/3 < 2π
5 0
3 years ago
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