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2 years ago

Guys pls help me im failing my grade and i really need help plsssss and dont do it for the points

Mathematics

1 answer:

Nataliya [291]2 years ago

6 0

Answer:

simplified: $\frac{3}{5}$

unsimplified: $\frac{12}{20}$

Step-by-step explanation:

Since the fractions have the same denominator, all we have to do is subtract the numbers in the numerator:

15 - 3 = 12

The difference of the two fractions is $\frac{12}{20}$ . To simplify this answer, divide both the numerator and the denominator by 4:

12/4 = 3

20/4 = 5

The difference in simplified form is $\frac{3}{5}$ .

I hope this helps. :)

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A rectangular athletic field is twice as long as it is wide. If the perimeter of the athletic field is 96 yards, what are its

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2 years ago

What is the ratio of melvins weight to the total weight of the two boys

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3 years ago

Cindy won the lottery, and she gave her family $1.000.00 less than of her winnings. If the jackpot was worth $35.000.00, how muc

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2 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

2 years ago