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Paraphin [41]
3 years ago
5

Find the parallel line to 5y-15x=-20 that goes through the point (3,-6).

Mathematics
1 answer:
Shtirlitz [24]3 years ago
8 0
5y-15x=-20
5y=15x-20
y=3x-20/3
Slope = 3

Slope Intercept Form
y-y1=m(x-x1)
Therefore y-(-6)=3(x-3)
y+6=3x-9
y=3x-15
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Step-by-step explanation:

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4 years ago
Question 1- Whats the derivative of: A) f(x)= 4 cos(x) + ln(x+1)<br> B) f(x)= sec(x) X tg(x)
Kryger [21]

The derivatives for this problem are given as follows:

a) f^{\prime}(x) = -4\sin{x} + \frac{1}{x + 1}

b) f^{\prime}(x) = \sec{x}\tan^{2}{x} + \sec^3{x}.

<h3>What is the derivative of the sum?</h3>

The derivative of the <u>sum is the sum of the derivatives</u>.

In this problem, the function is:

f(x) = 4\cos{x} + \ln{(x + 1)}

Using a derivative table for the derivatives of the cosine and the ln, the derivative of the function is:

f^{\prime}(x) = (4\cos{x})^{\prime} + (\ln{(x + 1)})^{\prime}

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What is the product rule?

The derivative of the product is given as follows:

(f(x) \times g(x))^{\prime} = f^{\prime}(x)g(x) + g^{\prime}(x)f(x)

In this problem, we have that:

  • f(x) = \sec{x}, f^{\prime}(x) = \sec{x}\tan{x}.
  • g(x) = \tan{x}, f^{\prime}(x) = \sec^2{x}.

Hence the derivative is:

f^{\prime}(x) = \sec{x}\tan^{2}{x} + \sec^3{x}.

More can be learned about derivatives at brainly.com/question/2256078

#SPJ1

4 0
2 years ago
What is the gradient of the graph shown?<br> Give your answer in its simplest form.
Minchanka [31]

Answer:

m=3/2

Step-by-step explanation:

Two points are shown

point 1: (0;-3). 4;3. -2-6

point 2: (2;0)

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m =  \frac{0 - ( - 3)}{2 - 0}

m =  \frac{3}{2}

6 0
3 years ago
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