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Dmitry [639]
3 years ago
6

When adding two rational numbers,if a pair of addends has opposite signs ,then the sum will have the sign of the addenda with th

e
Mathematics
1 answer:
____ [38]3 years ago
6 0
When adding two rational numbers,if a pair of addends has opposite signs ,then the sum will have the sign of the addenda with the greater  value among the two.

These examples can better illustrate this concept:
Example 1:
23 + (-3) = 20 (The greater among the two is positive and the result is also positive)

Example 2:
-23 + 3 = -20 (The greater among the two is negative and the result is also negative)
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If f(x) = 2x - 9, then f(3) = -3.<br><br> True<br> False
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True
work: f(3)=2(3)-9
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Please help me similar triangles make no sense to me
blondinia [14]

Answer:

Step-by-step explanation:

Compare the ratios of the triangles like so:

Bigger to smaller

\frac{15}{10}=\frac{12}{8}

This statement is true. Both fractions simplified are equal to 1.5. Since they have the same ratios, they should be similar.

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3 years ago
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3 years ago
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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
2 years ago
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