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AleksandrR [38]
3 years ago
11

Write the equation of the circle with center (5, 1) and radius r = 10

Mathematics
1 answer:
kherson [118]3 years ago
4 0
The standard equation for the circle is:

(x-h)^2+(y-k)^2=r^2, where (h,k) is the center of the circle and r is the radius.

We are given that the center is (5,1) and r=10 so

(x-5)^2+(y-1)^2=100
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Three-fourths of the tickets had been sold, and there were 420 tickets left. How many tickets were printed?
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Step-by-step explanation:

420 = 1/4 so 4/4 is 4 420s

420 x 4 = 1680

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2. Calculate an expression for dy/dx and d2y/dx2 in terms of t if the parametric pair is given as tan(x) = e^at and e^y = 1 + e^
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I assume a is a constant. If tan(x) = exp(at) (where exp(x) means eˣ), then differentiating both sides with respect to t gives

sec²(x) dx/dt = a exp(at)

Recall that

sec²(x) = 1 + tan²(x)

Then we have

(1 + tan²(x)) dx/dt = a exp(at)

(1 + exp(2at)) dx/dt = a exp(at)

dx/dt = a exp(at) / (1 + exp(2at))

If exp(y) = 1 + exp(2at), then differentiating with respect to t yields

exp(y) dy/dt = 2a exp(2at)

(1 + exp(2at)) dy/dt = 2a exp(2at)

dy/dt = 2a exp(2at) / (1 + exp(2at))

By the chain rule,

dy/dx = dy/dt • dt/dx = (dy/dt) / (dx/dt)

Then the first derivative is

dy/dx = (2a exp(2at) / (1 + exp(2at))) / (a exp(at) / (1 + exp(2at))

dy/dx = (2a exp(2at)) / (a exp(at))

dy/dx = 2 exp(at)

Since dy/dx is a function of t, if we differentiate dy/dx with respect to x, we have to use the chain rule again. Suppose we write

dy/dx = f(t)

By the chain rule, the derivative is

d²y/dx² = df/dx

d²y/dx² = df/dt • dt/dx

d²y/dx² = (df/dt) / (dx/dt)

d²y/dx² = 2a exp(at) / (a exp(at) / (1 + exp(2at)))

d²y/dx² = 2 (1 + exp(2at))

4 0
2 years ago
Harley paid $42 for a new pair of jeans. the store was having a sale dor 30%. What was the original price of the pair of jeans b
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Answer:

Original price = $60

Step-by-step explanation:

42 / 70 =0.6

0.6 x 100 = 60

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