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igor_vitrenko [27]
3 years ago
14

What is the result of rounding 8,888,888 to the nearest ten?

Mathematics
1 answer:
andreyandreev [35.5K]3 years ago
6 0

Answer:

8,888,890

Step-by-step explanation:

yup yup

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What is p equal to? Also if you could, can you tell me how you do it so I can do the rest by myself.
svp [43]

Answer:

p=-54

Step-by-step explanation:

- \frac{p}{6}  = 9 \\  \frac{ - p}{6}  = 9 \\ \frac{ - p}{6}  \times 6 = 9 \times 6 \\  - p = 54 \\ p =  - 54

6 0
3 years ago
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In a group of 58 learners at Nakambuda C.S, 34 learners have chosen the field of Agriculture. What fraction of learners will be
asambeis [7]

learners of Agriculture: 34/58

semplify it

divide by 2

34 > 17

58 > 29

they are prime numbers, so can't be semplify anymore

17/29 will do Agriculture

in percentage we have a proportion:

17 : 29 = x : 100

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4 0
3 years ago
Round to the nearest tenth
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Answer:

17.7

Step-by-step explanation:

a^{2} =29^{2} -23^{2} \\a=\sqrt{312} \\=17.6635

4 0
3 years ago
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Each side of a pentagon is 10 cm greater than the previous side. If the perimeter of this pentagon is 500 cm, find the lengths o
Vesna [10]

Answer: See explanation

Step-by-step explanation:

The perimeter of a pentagon is gotten through the summation of its five sides. Let the first side be represented by x. Since each side of a pentagon is 10 cm greater than the previous side, then the sides will be:

First side = x

Second side = x + 10

Third side = x + 10 + 10 = x + 20

Forth side = x + 30

Fifty side = x + 40

Therefore,

x + (x + 10) + (x + 20) + (x + 30) + (x + 40) = 500

5x + 100 = 500

5x = 500 - 100

5x = 400

x = 400/5

x = 80

Therefore, the lengths will be:

First side = x = 80cm

Second side = x + 10 = 80 + 10 = 90cm

Third side = x + 20 = 80 + 20 = 100cm

Forth side = x + 30 = 80 + 30 = 110cm

Fifty side = x + 40 = 80 + 40 = 120cm

8 0
3 years ago
Lusita earned $85 for 8 hours of babysitting. Avery earned $125 for 11 hours of babysitting. Are the rates equivalent
weeeeeb [17]
No, avery makes more then lusita
7 0
3 years ago
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