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Lapatulllka [165]
3 years ago
11

What is the solution to 3x2+7x−20=0

Mathematics
2 answers:
Misha Larkins [42]3 years ago
4 0

Answer:

Your answer is -4

Hope that this is helpful. Tap the crown button, Like & Follow me

den301095 [7]3 years ago
4 0

Answer:

x = 5/3

x = -4

-Evaluate the exponent

-Multiply the numbers

-Add the numbers

-Evaluate the square root

-Multiply the numbers

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6 3/4x

Step-by-step explanation:

If you follow PEMDAS (Parenthesis, Exponent, Multiply, Divide, Add, Subtract) then you do what is inside the parenthesis first.

3 1/2 divided by 2= 1 3/4

1 3/4 plus 5= 6 3/5

Then, you add the x and since there is no other variable like x, then you just leave it with 6 3/4.

Hope that helped! :)

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Four is the quotient of 24 divided by some number. What is the number?
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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)}\\\\\\
\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}\\\\\\
\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}}\\\\\\
\mathsf{2xy+x^2\cdot \dfrac{dy}{dx}-3-3y^2\cdot \dfrac{dy}{dx}=0}\\\\\\
\mathsf{x^2\cdot \dfrac{dy}{dx}-3y^2\cdot \dfrac{dy}{dx}=-\,2xy+3}\\\\\\
\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}}\\\\\\
\mathsf{\left.\dfrac{dy}{dx}\right|_{(-1,\,2)}=\dfrac{4+3}{1-12}}\\\\\\
\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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