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dimaraw [331]
3 years ago
9

a warehouse worker ships 9 boxes each day. Every box must contain 3 shipping labels .How many shipping labels does the worker ne

ed each day?
Mathematics
2 answers:
aev [14]3 years ago
7 0

9 (boxes a day) x 3 (shipping labels per box)= 27 labels per day

Kisachek [45]3 years ago
6 0

Answer: 27

Step-by-step explanation:

Given : The number of boxes warehouse worker ships = 9

The number of shipping labels contained in box = 3

Then the number of shipping labels the worker need each day is given by :-

9\times3=27

Hence, the number of shipping labels the worker need each day =27

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Answer:

91.125 or 729/8

Step-by-step explanation:

take 4.5^3 to get 91.125

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Find the circumference of a circle with a radius of 2.4 m
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15.07964

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First tour is 4 child tickets and 8 adult tickets for 128 dollars second tour is 6 adult tickets for 72 dollars what is the pric
V125BC [204]

Answer:the price of one child ticket is $8

the price of one adult ticket is $12

Step-by-step explanation:

Let x represent the price of one child ticket.

Let y represent the price of one adult ticket.

First tour is 4 child tickets and 8 adult tickets for 128 dollars. This means that

4x + 8y = 128 - - - - - - - - - - 1

Second tour is 6 adult tickets for 72 dollars. This means that

6y = 72

y = 72/6 = $12

Substituting y = 12 into equation 1, it becomes

4x + 8 × 12= 128

4x + 96 = 128

4x = 128 - 96

4x = 32

x = 32/4 = $8

6 0
3 years ago
Find general solutions of the differential equation. Primes denote derivatives with respect to x.
zepelin [54]

Answer:

\mathbf{3x^2y^3+2xy^4=C}

Step-by-step explanation:

From the differential equation given:

6xy^3 +2y ^4 +(9x^2y^2+8xy^3) y' = 0

The equation above can be re-written as:

6xy^3 +2y^4 +(9x^2y^2+8xy^3)\dfrac{dy}{dx}=0

(6xy^3 +2y^4)dx +(9x^2y^2+8xy^3)dy=0

Let assume that if function M(x,y) and N(x,y) are continuous and have continuous first-order partial derivatives.

Then;

M(x,y) dx + N (x,y)dy = 0; this is exact in R if and only if:

\dfrac{{\partial M }}{{\partial y }}= \dfrac{\partial N}{\partial  x}}} \ \ \text{at each point of R}

relating with equation M(x,y)dx + N(x,y) dy = 0

Then;

M(x,y) = 6xy^3 +2y^4\  and \ N(x,y) = 9x^2 y^2 +8xy^3

So;

\dfrac{\partial M}{\partial y }= 18xy^3 +8y^3

       \dfrac{\partial N}{\partial y }

Let's Integrate \dfrac{\partial F}{\partial x}= M(x,y) with respect to x

Then;

F(x,y) = \int (6xy^3 +2y^4) \ dx

F(x,y) = 3x^2 y^3 +2xy^4 +g(y)

Now, we will have to differentiate the above equation with respect to y and set \dfrac{\partial F}{\partial x}= N(x,y); we have:

\dfrac{\partial F}{\partial y} = \dfrac{\partial}{\partial y } (3x^2y^3+2xy^4+g(y)) \\ \\ = 9x^2y^2 +8xy^3 +g'(y) \\ \\ 9x^2y^2 +8xy^3 +g'(y) =9x^2y^2 +8xy^3 \\ \\ g'(y) = 0  \\ \\ g(y) = C_1

Hence, F(x,y) = 3x^2y^3 +2xy^4 +g(y)  \\ \\ F(x,y) = 3x^2y^3 + 2xy^4 +C_1

Finally; the general solution to the equation is:

\mathbf{3x^2y^3+2xy^4=C}

5 0
3 years ago
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