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sladkih [1.3K]
2 years ago
14

2. Twenty machines can embroider 1000 sweaters in 5 days.

Mathematics
1 answer:
Sonja [21]2 years ago
5 0

Answer:

a) 20 days

b) 50 machines

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What is the solution to this inequality?
ioda
C. we multiply the equation by 3
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3 years ago
Help me out please! everyday !
Lady bird [3.3K]

Answer:

V = 904.32 ft^3

Step-by-step explanation:

The volume of a sphere is given by

V = 4/3 pi r^3

The radius is 6 and pi = 3.14

V = 4/3 ( 3.14) (6)^3

V = 904.31999 ft^3

Rounding to the hundredths place

V = 904.32 ft^3

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3 years ago
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Vupti vvo 1.5(A): Solving Linear Equatior<br> your work.<br> 4n-10 =12
Lubov Fominskaja [6]

Answer:

n = 5.5

Step-by-step explanation:

4n - 10 = 12

       4n = 12 + 10

       4n = 22

          n = 22/4

          n = 5.5

5 0
3 years ago
Can you pls help me i need it?
Elina [12.6K]

Answer:

$114.75

Step-by-step explanation:

You have to multiply the hours by the wage. Looking at the time that the worker was in during the morning, it was a total of 4 hours. Since the wage is $13.50/hr, we would multiply 13.50 by 4.

13.50 * 4 = 54.00

So, now we have to add together the total hours in the afternoon. If we count the time, we get 4 1/2 hours. So, now we multiply 13.50 by 4.5.

13.50 * 4.5 = 60.75

Now, to find the total pay for that day, we add both the morning and the afternoon pay together.

54.00 + 60.75 = 114.75

Therefore, the pay for this day is $114.75.

6 0
2 years ago
Consider the differential equation
Gennadij [26K]

The first solution is quadratic, so its derivative y' on the left side is linear. But the right side would be a polynomial of degree greater than 1, so this is not the correct choice.

The third solution has a similar issue. The derivative of √(x² + 1) will be another expression involving √(x² + 1) on the left side, yet on the right we have y² = x² + 1, so that the entire right side is a polynomial. But polynomials are free of rational powers, so this solution can't work.

This leaves us with the second choice. Recall that

1 + tan²(x) = sec²(x)

and the derivative of tangent,

(tan(x))' = sec²(x)

Also notice that the ODE contains 1 + y². Now, if y = tan(x³/3 + 2), then

y' = sec²(x³/3 + 2) • x²

and substituting y and y' into the ODE gives

sec²(x³/3 + 2) • x² = x² (1 + tan²(x³/3 + 2))

x² sec²(x³/3 + 2) = x² sec²(x³/3 + 2)

which is an identity.

So the solution is y = tan(x³/3 + 2).

4 0
2 years ago
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