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alukav5142 [94]
2 years ago
9

Solve for x to the nearest degree.

Mathematics
1 answer:
VikaD [51]2 years ago
5 0

Answer:

Using Pythagoras Theorem, squareroot 15^2 - 5^2.

After you found the opposite length, use the cosine rule to find angle x.

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Given h(x) = -2- 4, find h(-6)
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Answer:

h(-6) = 2

Step-by-step explanation:

-x - 4 =

-(-6) - 4 = 2

6-4=2

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On a piece of paper, graph y<-2/5x+1, then determine which answer choice matches the graph you drew
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What is the value of \dfrac{d}{dx}\left(\dfrac{2x+3}{3x^2-4}\right) dx d ​ ( 3x 2 −4 2x+3 ​ )start fraction, d, divided by, d, x
stellarik [79]

Answer:

4.

Step-by-step explanation:

We are asked to find the value of expression \frac{d}{dx}(\frac{2x+3}{3x^2-4}) at x=-1.

First of all, we will find the derivative of the given expression using "Quotient Rule of Derivatives" as shown below:

(\frac{f(x)}{g(x)})'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{(g(x))^2}

\frac{d}{dx}(\frac{2x+3}{3x^2-4})

\frac{\frac{d}{dx}(2x+3)*(3x^2-4)-(2x+3)*\frac{d}{dx}(3x^2-4)}{(3x^2-4)^2}

\frac{2*(3x^2-4)-(2x+3)*(6x)}{(3x^2-4)^2}

\frac{6x^2-8-12x^2-18x}{(3x^2-4)^2}

\frac{-6x^2-18x-8}{(3x^2-4)^2}

Therefore, our required derivative is \frac{-6x^2-18x-8}{(3x^2-4)^2}.

Now, we will substitute x=-1 in our derivative to find the required value as:

\frac{-6(-1)^2-18(-1)-8}{(3(-1)^2-4)^2}

\frac{-6(1)+18-8}{(3(1)-4)^2}

\frac{-6+18-8}{(3-4)^2}

\frac{4}{(-1)^2}

\frac{4}{1}

4

Therefore, the value of expression \frac{d}{dx}(\frac{2x+3}{3x^2-4}) at x=-1 is 4.

6 0
3 years ago
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