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lina2011 [118]
3 years ago
11

Find the parametric equations for the line that is tangent to the given curve at the given parameter (3cost)i (t^4 -3sint)j

Mathematics
1 answer:
ale4655 [162]3 years ago
5 0

Answer:

Following are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

In the given equation, when the point t=0

So,

\to r(0) = (3 \cos 0)i + (0^4 - 6 \sin 0)j + (2e^{3\times 0})k)

           = (3 \times 1)i + (0 - 0)j + (2e^{0})k)\\\\ = 3i +  0j + (2 \times 1)k)\\\\ = 3i +  0j + 2k \\

The value of the coordinates are 3, 0, 2 . so, the equation of the line is:

\to \frac{(x-3)}{3 \cos \ t} = \frac{(y-0)}{(t^{4}-6 \sin \ t)} = \frac{(z-2)}{2e^{3t}}=k

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If f(x)=x+8 and g(x)=-4x-3 find (f+g)(x)
nydimaria [60]

Answer:

for this case we have the following functions:

f (x) = x + 8

g (x) = -4x - 3

Subtracting the functions we have:

(f - g) (x) = f (x) - g (x)

(f - g) (x) = (x + 8) - (-4x - 3)

Rewriting:

(f - g) (x) = x + 8 + 4x + 3

(f - g) (x) = 5x + 11

Answer:

D. (f - g) (x) = 5x + 11

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3 years ago
What is the solution to this system of equations?<br> x = 12 − y<br> 2x + 3y = 29
Elenna [48]
Let’s arrange the fist equation
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2x+3y=29

Multiply equation 1 by 2
So it will be

2x+2y=24

Subtract eq 1 from eq 2

2x+3y=29
2x+2y=24
—————-
Y =5

Put value of y in eq 1

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S.S= (7,5)
8 0
2 years ago
Read 2 more answers
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Answer

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Step-by-step explanation:

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3 years ago
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uysha [10]
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3 years ago
Use Natural Logarithms to solve the equation
Rus_ich [418]

Answer:

A) 0.106

Step-by-step explanation:

8e^3x + 4 = 15

Subtract 4 from each side

8e^3x + 4-4 = 15-4

8e^3x  = 11  

Divide each side by 8

8/8e^3x = 11/8

e^(3x) = 11/8

Take the natural log of each side

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3x = ln(11/8)

Divide by 3

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To the nearest thousandth

x = .106

7 0
3 years ago
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