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3 years ago

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]

Answer:

The Answer is B.

Step-by-step explanation:

I took the quiz on Edge.

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3 years ago

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How can you tell if a line goes through the origin

photoshop1234 [79]

If it's connected to it.

6 0

3 years ago

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

5 0

4 years ago

Find general solutions of the differential equation. Primes denote derivatives with respect toÂ 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