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tensa zangetsu [6.8K]
3 years ago
5

Find the slope of the line that contains the points (6,3) and (12,0).

Mathematics
2 answers:
777dan777 [17]3 years ago
6 0
M= y2-y1/x2-x1
3-0/6-12= 3/-6= -1/2
(Slope) m=-1/2
Yuki888 [10]3 years ago
6 0
Y=-.5x+6 is your slope intercept equation.
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Find dy/dx for y= x^3 ln (cot x)
ICE Princess25 [194]
<h3>Answer</h3>

  \dfrac{dy}{dx} = 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)

<h3>Explanation</h3>

By the product rule (d/dx)(f(x)g(x)) = f(x)g'(x) + g(x)f'(x), we have

  \begin{aligned}\frac{dy}{dx} &= \left(x^3 \ln (\cot x) \right)' \\&= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \end{aligned}

By the chain rule:

  \begin{aligned}\big(\ln (\cot x)\big)' &= \dfrac{1}{\cot x} \cdot (\cot x)' \\ &= \dfrac{1}{\cot x} \cdot -\csc^2 x\\&= -\tan (x) \csc^2(x) \\&= - \frac{\sin x}{\cos x} \cdot \frac{1}{\sin^2 x} = - \frac{1}{\cos x} \cdot \frac{1}{\sin x} \\&= -\csc(x)\sec(x)\end{aligned}

By the power rule:

  (x^3)' = 3x^2

thus

  \begin{aligned}\frac{dy}{dx} &= x^3\big(\ln (\cot x)\big)' + \ln (\cot x) \cdot \left(x^3\right)' \\&= x^3\big( -\csc(x)\sec(x) \big) + \ln(\cot x) \cdot (3x^2) \\&= -x^3 \csc(x)\sec(x) + 3x^2 \ln(\cot x) \\&= 3x^2 \ln(\cot x)-x^3 \csc(x)\sec(x)\end{aligned}

Nothing to do to simplify any further, other than factoring out x^2.

4 0
4 years ago
PLEASE HELP!!!!! What are the values of a and b? -7(x+3)=ax+b
Artist 52 [7]

Step-by-step explanation:

Hope dis helps you......

3 0
3 years ago
Multiply each equation by a constant that would help to eliminate the y terms.
Zinaida [17]

Answer:

Therefore the required resulting equation is

6x-15y=-63\\-15x+15y=90

6 x minus 15 y = negative 63.

Negative 15 x + 15 y = 90

Step-by-step explanation:

Given:

2x-5y=-21         ......................Equation ( 1 )

3x-3y=-18         ......................Equation ( 2 )

To Find:

Expression after multiplying to eliminate y term,

Solution:

So to eliminate 'y' term we need to multiply equation 1 by a constant 3 and equation 2 by a constant -5,  such that equations  becomes

3(2x-5y)=3\times -21\\\\6x-15y=-63 .....( 1 )

-5(3x-3y)=-5\times -18\\\\-15x+15y=90  .....( 2 )

so now by adding new equation one and two we can eliminate y term that means -15y and +15y will get cancel,

Therefore the required resulting equation is

6x-15y=-63\\-15x+15y=90

6 x minus 15 y = negative 63.

Negative 15 x + 15 y = 90

3 0
3 years ago
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In certain economies, the government operates as the central authority, guiding the economy and controlling (and owning) many of
Andrej [43]

Answer:

b

Step-by-step explanation:

in certain economies, the government operates as the central authority, guiding the economy and controlling (and owning) many of the businesses. Instead of being produced to make a profit, goods and services are produced for uses planned by the government. What type of economy does this describe? A. gift economy B. market economy C. socialist economy D. laissez-faire economy

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3 years ago
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Find y when x=9 if y varies directly as the square of x, and y=100 when x=5.
sukhopar [10]
If y varies directly as the square of x, that means that y=k*x^2. Plugging y=100 and x=5 into it, we get 100=k*5^2=k*25. Dividing by both sides, we get k=100/25=4. Going back to the original equation, we now know that y=4*x^2. Plugging 9=x in, we get 4*9^2=4*81=324=y
7 0
3 years ago
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