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lesya692 [45]
3 years ago
15

PLZ ONLY TYPE IN SSS , ASA, SAS, AAS, HL) IF the triangles are not congruent, type in NEI

Mathematics
1 answer:
yanalaym [24]3 years ago
4 0
Sas sss is the correct answer
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Helen got a $7 reduction in the price of a pair of jeans that are normally $32. What is the approximate percent decrease in the
Vesnalui [34]

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The Answer is B.

Step-by-step explanation:

I took the quiz on Edge.

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How can you tell if a line goes through the origin
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If it's connected to it.
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PLEASE HELP ASAP FIRST TO ANSWER CORRECTLY GETS BRAINLYIST (question in the pic)
Simora [160]

C i believe because they both can break it up those ways even if they might not be the most effective

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4 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
The equation of the line in the graph is y=___ x+=_____<br> (-1,-1)
il63 [147K]

Answer:

y= 1x-1 or y= x-1

Step-by-step explanation:

Make sure we know the equation y = mx +b

Remember M= slope, B= Y-intercept

We can see that the slope (rise over run)  is 1 so we plug in the 1 for M

Y= 1x +b

Now we have to look for the Y-intercept which is -1

making our equation now: Y= 1x -1

3 0
3 years ago
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