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Rama09 [41]
3 years ago
5

R=8cos(theta)-10sin(theta) to cartesian coordinates

Mathematics
1 answer:
dsp733 years ago
8 0
Convert polar equation (R, Ф):

(1) R= 8cos Ф - 10sin Ф into Cartesian (x , y)
We know that:
x = R.cos Ф  → cos Ф = x/R , and
y = R.sin Ф →  sin Ф = y/R
and that x² +y² = R²

Replace in 
R= 8cos Ф - 10sin Ф , cos Ф and sin Ф by the related x and y:

R = 8(x/R) - 10(y/R)
Multiply both sides by R:

R² = 8R(x/R) - 10R(y/r) ↔ R² = 8x - 10y , but we have also R² = x² + y². hence: 8x -10y = x² + y²
OR x² + y² - 8x + 10y = 0
We can continue in completing th squares of (x² - 8x + ?) and (y² +10y + ?)

(x² - 8x + ?) = (x - 4)² - 16

and  (y² +10y + ?) = (y +5)² - 25
Final Equation:
(x - 4)² - 16 + (y +5)² - 25 →→→(x - 4)² + (y +5)² = 41 
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Hope this helps you!

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Assume that during each second, a job arrives at a webserver with probability 0.03. Use the Poisson distribution to estimate the
Savatey [412]

Answer:

The probability that at lest one job will be missed in 57 second is=1- e^{-1.71} =0.819134

Step-by-step explanation:

Poisson distribution:

A discrete random variable X having the enumerable set {0,1,2,....} as the spectrum, is said to be Poisson distribution.

P(X=x)=\frac{e^{-\lambda t}({-\lambda t})^x}{x!}  for x=0,1,2...

λ is the average per unit time

Given that, a job arrives at a web server with the probability 0.03.

Here λ=0.03, t=57 second.

The probability that at lest one job will be missed in 57 second is

=P(X≥1)

=1- P(X<1)

=1- P(X=0)

=1-\frac{e^{-1.71}(1.71)^0}{0!}

=1- e^{-1.71}

=0.819134

6 0
3 years ago
Factor 2x 2 - 10x - 12.
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2(x+1)(x- 6) this is probably correct
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What are the domain and range of the function? F(x)=-3/5x^3
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Answer:

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Range: All real numbers, excluding zero      (-∞, 0) ∪ (0, ∞)

Step-by-step explanation:

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