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kenny6666 [7]
4 years ago
14

Can someone simplify 2y-3x^2+6x^2-3y ?

Mathematics
2 answers:
ahrayia [7]4 years ago
8 0
2y - 3x^2 + 6x^2 - 3y
Bring like terms together:-
6x^2 - 3x^2 + 2y - 3y
= 3x^2 - y   <-----  Answer

statuscvo [17]4 years ago
5 0
2y-3x^2+6x^2-3y  combine like terms

3x^2-y
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The value of m varies inversely as the square of n. When n = 3 m = 6.
ss7ja [257]

Answer:

B.]2

Step-by-step explanation:

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4 years ago
A dairy farm had 9 pastures and 630 cows. The same number of cows are placed in each pasture. How many cows are in each pasture?
rjkz [21]

Answer:


Step-by-step explanation:

if a dairy farm has 630 cows and the same number of cows are placed in each pasture the total number of cows in each pasture is 630

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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
When is it easy to show if a set is closed or not closed?
Pani-rosa [81]

Answer:i think its if the dots are open they arent closed but if they are filled in its closed same with () and [] if its (x,y) then its not able to be used but if its [x,y] then it is able to be used

Step-by-step explanation:

3 0
3 years ago
Never was given anything to help solve they just assume you know it
makkiz [27]

Answer: 0.4575

Step-by-step explanation:

.915/2

4 0
3 years ago
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