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devlian [24]
3 years ago
6

The class president is organizing a class trip to a nearby amusement park for 314 students. The regular price is $35 per ticket.

However, some students can receive a discount due to volunteer community service work that they took part in on Saturdays. The students who are eligible for the discount will pay $21.50. The total ticket cost for the class trip will be $10,072. How many students are eligible for the discount?
Mathematics
1 answer:
salantis [7]3 years ago
7 0

Answer:

287 students were eligible for the discount :)

Step-by-step explanation:

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Which best describes the transformation that occurs from the graph of f(x)=x^2 to g(x)=(x-2)^2+3?? ​
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answer right 2, up 3

any function with the following form is a transformation from f(x) = x²

g(x) = (x – a)² + b

were a moves the function to the right when a is a positive number and to the left when its a negative number, and b moves the function up when b is positive and down when its negative.

then for a=2 positive and b=3 positive, we have

right 2, up 3

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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}

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2 years ago
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andrey2020 [161]

Answer:

The median is 5, mode is 5, and range is 12

Step-by-step explanation:

6 0
2 years ago
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