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valina [46]
3 years ago
14

Find the number of steel ball bearings, each of diameter 0.7 cm, that can be made from 1kg of steel, given that 1cm cubed of ste

el weighs 7.8 g​
Mathematics
1 answer:
dimulka [17.4K]3 years ago
7 0

Answer:

Ball Bearing Volume = (4 / 3) * PI * radius ^3

Volume = (4 / 3) * 3.14159265 * (.35)^3

Volume = (4 / 3) * 3.14159265 * 0.042875

Volume = 0.17959438 cc

1 cc of steel = 7.8 g

Volume of 1 kg of steel = 1,000 cc / 7.8 g per cc

1 kg = 128.205 cc

Number of ball bearings we can make = 128.205 / 0.17959438

Equals 713.86

Step-by-step explanation:

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Determine the values of k and p so that the solution is "All Reals" when solving for y:
Angelina_Jolie [31]

Answer:

The value of k is  \pm 9 and The value of p is - 23  

Step-by-step explanation:

Given equation as :

81 y - 17  = k² y + 6 + p

The equation has real solution ,

Now for real solution the equation have no solution

∵ The equation has no solution the ,

The coefficient of y must be equal

I.e 81 = k²

Or, k = \sqrt{81}

∴   k = \pm 9

Again , If the coefficient of y is same then the equation is written as

6 + p = - 17

or, p = - 17 - 6

or, p = - 23

Hence The value of k is  \pm 9 and The value of p is - 23   Answer

6 0
3 years ago
there are 3 method to show that 40 is the sum of two prime. among all the prime , find the difference between the two largest pr
Maurinko [17]

Answer:

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3 years ago
Read 2 more answers
5 yrs ago, Nuri was thrice as old as Sonu. 10 yrs later, Nuri will be twice as old Sonu. How old are Nuri n Sonu?​
Papessa [141]

Answer:

Answer will be 50

Step-by-step explanation:

Let us suppose, present age of Nuri be ‘x’ years and present age of Sonu be ‘y’ years.

Now, it is given that five years ago, Nuri was thrice old as Sonu. Hence,

Five years ago,

Nuri’s age = x-5 years

Sonu’s age = y-5 years

And relation between ages can be given as

Nuri’s age = 3×sonu’s age or

x-5 = 3(y-5)

x-5 = 3y-15

x-3y+10 = 0 ………..(i)

Another relation is given in the problem that ten years later, Nuri is twice as old as Sonu.

So, ten years ago,

Nuri’s Age = x+10

Sonu’s Age = y+10

And relation between ages can be written as

x+10 = 2(y+10)

x+10 = 2y+20

x-2y-10 = 0 …………..(ii)

Now we can solve the equation (i) and (ii) to get values of x and ‘y’ or present ages of Nuri and Sonu.

Value of ‘x’ from equation (i) be

x = 3y-10 ……….(iii)

Putting value of ‘x’ from equation (iii) in equation (ii) we get,

3y-10-2y-10 = 0

y = 20

Now, from equation (iii) value of ’x’ can be given as,

x= 3(20)-10

x = 50

Hence, the present ages of Nuri and Sonu are 50 years and 20 years respectively.

7 0
2 years ago
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Answer:

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4 0
3 years ago
Factor the expression completely.
RSB [31]

 -9.75 + 3.25x

= -3.25(3) + 3.25(x)

= -3.25(3)  -  -3.25(x)     <em>notice they have the same factor of -3.25</em>

= -3.25(3 - x)

Answer: A


3 0
3 years ago
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