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gulaghasi [49]
3 years ago
6

Find equation of the circle in standard form for the given center (h,k) and radius (r) : (h,k)=( -3/5 , -4/5 ), r=1

Mathematics
1 answer:
svp [43]3 years ago
4 0
The center is at (h,k) so we have the folowing.
(x - (-\frac{3}{5}))^2 + (x - (-\frac{4}{5}))^2 
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Answer:

C

Step-by-step explanation:

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Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = e−5t
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Answer:

x = 1 - 5t

y = t

z = 1 - 5t

Step-by-step explanation:

For the equation of a line, we need a point and a direction vector. We are given a point (1, 0, 1).

Since the line is suppose to be a tangent to the given curve at the point (1, 0, 1), we need to find a tangent vector for which the curve passes through that point.

We have x = e^(-5t)cos5t

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at t = 0, y = 0

z = e^(-5t)

at t = 1, z = e^(-5)

at t = 0, z = 1

Clearly, the only parameter value for which the curve passes through the point (1, 0, 1) is t = 0.

In vector notation, the curve

r(t) = xi + yj + zk

= e^(-5t)cos5t i + e^(-5t)sin5t j + e^(-5t) k

r'(t) = [-5e^(-5t)cos5t - e^(-5t)sin5t] i +[e^(-5t)cos5t - 5e^(-5t)sin5t] j - 5e^(-5t) k

r'(0) = -5i + j - 5k

is a vector tangent at the point.

We get the parametric equation from this.

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y = y(0) + ty'(0)

= t

z = z(0) + tz'(0)

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The answers below ._.

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Question 1 (1 point)
ratelena [41]
I have no idea I’m sorry
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3 years ago
After reading a book foe English class, 100 students were asked whether or not they enjoyed it. Nine twenty-fifths of the class
Katena32 [7]

Answer:

The number of students who liked the book was 64.

Step-by-step explanation:

After reading a book for English class, 100 students were asked whether or not they enjoyed it.

Now, given that nine twenty-fifths of the class did not like the book.

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Therefore, the number of students who liked the book was (100 \times \frac{16}{25}) = 64. (Answer)

7 0
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