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Sedaia [141]
3 years ago
13

Which of the following points are in the first quadrant of the xy-plabe? check all that apply

Mathematics
2 answers:
Serga [27]3 years ago
8 0

Answer: B and D because for the 1st quadrant both x and y need to be positive

Step-by-step explanation:

Readme [11.4K]3 years ago
7 0

B. (2,2)

D. (4,8)

The format for 1st quadrant is: (+,+).

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Write 100 times 100as multiplying 10 to a power by 10 to a power?<br>​
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10^2 × 10^2

Step-by-step explanation:

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3 years ago
a The function f(x) = |x - 3| can be used to determine how far a number is away from the number 3 on the number line. Find and i
jasenka [17]

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Please add the given function values. Or add an image.

Step-by-step explanation:

7 0
3 years ago
If x^2y-3x=y^3-3, then at the point (-1,2), (dy/dx)?
zavuch27 [327]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2866883

_______________


          dy
Find  ——  for an implicit function:
          dx


x²y – 3x = y³ – 3


First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:

\mathsf{\dfrac{d}{dx}(x^2 y-3x)=\dfrac{d}{dx}(y^3-3)}\\\\\\&#10;\mathsf{\dfrac{d}{dx}(x^2 y)-3\,\dfrac{d}{dx}(x)=\dfrac{d}{dx}(y^3)-\dfrac{d}{dx}(3)}


Applying the product rule for the first term at the left-hand side:

\mathsf{\left[\dfrac{d}{dx}(x^2)\cdot y+x^2\cdot \dfrac{d}{dx}(y)\right]-3\cdot 1=3y^2\cdot \dfrac{dy}{dx}-0}\\\\\\&#10;\mathsf{\left[2x\cdot y+x^2\cdot \dfrac{dy}{dx}\right]-3=3y^2\cdot \dfrac{dy}{dx}}


                        dy
Now, isolate  ——  in the equation above:
                        dx

\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3=3y^2\cdot \dfrac{dy}{dx}}\\\\\\&#10;\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3-3y^2\cdot \dfrac{dy}{dx}=0}\\\\\\&#10;\mathsf{x^2\cdot \dfrac{dy}{dx}-3y^2\cdot \dfrac{dy}{dx}=-\,2xy+3}\\\\\\&#10;\mathsf{(x^2-3y^2)\cdot \dfrac{dy}{dx}=-\,2xy+3}


\mathsf{\dfrac{dy}{dx}=\dfrac{-\,2xy+3}{x^2-3y^2}\qquad\quad for~~x^2-3y^2\ne 0}


Compute the derivative value at the point (– 1, 2):

x = – 1   and   y = 2


\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{-\,2\cdot (-1)\cdot 2+3}{(-1)^2-3\cdot 2^2}}\\\\\\&#10;\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{4+3}{1-12}}\\\\\\&#10;\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{7}{-11}}\\\\\\\\ \therefore~~\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=-\,\dfrac{7}{11}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)


Tags:  <em>implicit function derivative implicit differentiation chain product rule differential integral calculus</em>

6 0
3 years ago
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