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PSYCHO15rus [73]
3 years ago
7

Help with this please

Mathematics
1 answer:
AfilCa [17]3 years ago
3 0

Answer:

10

Step-by-step explanation:

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Convert each percent into a fraction and decimal:<br> 133 1/2%<br> 87 1/3%<br> 0.01%<br> 8 1/3%
Elis [28]

9514 1404 393

Answer:

  • 267/200 = 1.335
  • 131/150 = 0.8733...
  • 1/10,000 = 0.0001
  • 1/12 = 0.0833...

Step-by-step explanation:

It might be helpful to think of the % symbol as being interchangeable with /100.

133 1/2% = (133 1/2)/100 = 267/200 = 1.335

87 1/3% = (87 1/3)/100 = 262/300 = 131/150 = 0.8733333... (repeating 3)

0.01% = 0.01/100 = 1/10000 = 0.0001

8 1/3% = (8 1/3)/100 = 25/300 = 1/12 = 0.0833333... (repeating 3)

6 0
3 years ago
George is building a large model airplane in his workshop. If the door to his workshop is 3 ft wide and 6 1/2 ft high and the ai
Vitek1552 [10]
8282866-(9288)[81818]
7 0
4 years ago
Which of the following integrals represents the area of the region bounded in the first quadrant by x = pi/ 4 and the functions
Kaylis [27]

Answer:

option B is true.

Step-by-step explanation:

We are given that two functions

f(x)=sec^2x and g(x)=sin x and a line x =\frac{\pi}{4}

We have to find the area of the region bounded in the first quadrant by x={\pi}{4} and two functions

We know that the area bounded by two functions

=Integration of region(Upper curve- lower curve)

Therefore, function of sec square x is upper curve and function of sin x is lower function

Therefore, limit of x changing from 0 to \frac{\pi}{4}

Hence, the area of the region bounded in the first quadrant and two functions is given by

=\int_{0}^{\frac{\pi}{4}} (sec^2x-sinx) dx

Therefore, option B is true.

7 0
3 years ago
To find 27x9 I find 27x10 and remove 1 group of 27
solniwko [45]

Answer:

yes because 10 groups 27 taking 1 group of 27 is now 9 groups of 27

10*27-27=243 is equal to 9*27=243

5 0
3 years ago
Decompose the number 419 in 10 models
Papessa [141]
100
100
100
100
3
2
5
3
3
3

I think this is what you're asking.. If not.. I'm sorry.

5 0
3 years ago
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