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3 years ago

11

Team A is located on the south bank of a river. Teams B and C are located on the north bank. The angle formed from Team C to Tea

m A is S 20o W. Team C is 45 feet due west of Team B and 25 feet from Team A. How far is Team A from Team B?
23.1 ft

47.6 ft

58.5 ft

43.4 ft

1 answer:

creativ13 [48]3 years ago

6 0

The correct answer is 58.5

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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)}\\\\\\
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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>

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3 years ago

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shepuryov [24]

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Step-by-step explanation:

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2 years ago

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Please help me on this

Vadim26 [7]

Answer:

The answer is expression 4㏒w(x² - 6) - (1/3)㏒w(x² + 8) ⇒ 3rd answer

Step-by-step explanation:

* Lets revise some rules of the logarithmic functions

- log(a^n) = n log(a)

- log(a) + log(b) = log(ab) ⇒ vice versa

- log(a) - log(b) = log(a/b) ⇒ vice versa

* Lets solve the problem

- The expression is

$log_{w}\frac{(x^{2}-6)^{4}}{\sqrt[3]{x^{2}+8}}$

∵ log(a/b) = log(a) - log(b)

∴ $log_{w}(x^{2}-6)^{4}-log_{w}\sqrt[3]{x^{2}+8}$

∵ ∛(x² + 8) can be written as (x² + 8)^(1/3)

∵ log(a^n) = n log(a)

∴ $log_{w}(x^{2}-6)^{4}=4log_{w}(x^{2}-6)$

∴ $log_{w}\sqrt[3]{x^{2}+8}=\frac{1}{3} log_{w} (x^{2}+8)$

∴ $4log_{w}(x^{2}-6)-\frac{1}{3}log_{w}(x^{2}+8)$

* The answer is expression 4㏒w(x² - 6) - (1/3)㏒w(x² + 8)

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3 years ago