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Sergio039 [100]
3 years ago
13

In a toy factory 200 wooden soils cylinders 7 cm long and 35 mm in diameter have to be painted. What is the total surface area i

n cm that needs to be painted
Mathematics
1 answer:
emmainna [20.7K]3 years ago
6 0
By applying 2×pie×radius (radius+height)
This answer comes 1925000mm^3
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Liono4ka [1.6K]

Distribute the 2 first.

f(n) = 2n - 2 + 3n

Combine like terms.

f(n) = 5n - 2

8 0
3 years ago
Please help im stressing :((((<br> (what is Danielle mistake in this problem )
statuscvo [17]

Answer:

Daniel's Mistake was that she didnt add all the sides

Step-by-step explanation:

She only added the right side of the triangle, and the right and bottom sides of the square. She is missing the left of the triangle, which is another 6x-4, and the left side of the square, which is 12x +3.

Using these sides the answer should look like this,

set up your problem

1) (6x-4)*2 + (12x+3)*2 + (14x+13)

multiply your 2's in

2) (12x-8) + (24x+6) + (14x+13)

simplify

3) 50x + 11 is your answer

7 0
3 years ago
Solve the pair of simultaneous equations and leave the answer as a fraction in
Elena-2011 [213]

Answer:

x = 11/3 = 3 2/3

y = 13/3 = 4 1/3

Step-by-step explanation:

Here, we want to solve the system of equations simultaneously

x + y = 8

2x -3 = y

From the second equation, we have an equation for y

we can simply proceed to substitute this into the first equation

x + 2x - 3 = 8

3x - 3 = 8

3x = 8 + 3

3x = 11

x = 11/3

Recall;

y = 2x - 3

y = 2(11/3) - 3

y = 22/3 - 3

y = (22-3(3))/3

y = (22-9)/3 = 13/3

6 0
3 years ago
What is the area of a rectangle with vertices at (6, −3) ​, ​ (3, −6) ​ , (−1, −2), and (2, 1) ?
Bess [88]

Answer:

-9.Try to caculate it, My friend

5 0
3 years ago
Question A volleyball team sold raffle tickets to raise money for the upcoming season. They sold three different types of ticket
m_a_m_a [10]

Answer:

24 premium tickets were sold.

Step-by-step explanation:

Let :

Deluxe ticket = x

Regular tickets = x + 78

Premium tickets = y

x + (x + 78) + y = 208

4x + 2(x+78) + 10y = 714

2x + y = 208 - 78

4x + 2x + 156 + 10y = 714

2x + y = 130 - - - - - (1)

6x + 10y = 558 - - - - (2)

Now we can solve the simultaneous equation using elimination method :

From (1)

y = 130 - 2x

Put y = 130 - 2x in (2)

6x + 10(130 - 2x) = 558

6x + 1300 - 20x = 558

- 14x = 558 - 1300

-14x = - 742

x = 742 / 14

x = 53

Put x = 53 in y = 130 - 2x

y = 130 - 2(53)

y = 130 - 106

y = 24

3 0
3 years ago
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