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3 years ago

write a polar equation of a conic with the focus at the origin and the given data.hyperbola, eccentricity 3.5, directrix y

Mathematics

1 answer:

Colt1911 [192]3 years ago

7 0

Complete Question is;

Write a polar equation of a conic with the focus at the origin and the given data. hyperbola, eccentricity 3.5, directrix y =2

Answer:

r = 14/(2 + 7(sinθ))

Step-by-step explanation:

We are given;

Eccentricity;e = 3.5

Directrix; y = 2

This means that d = 2

We are told that the focus is at the origin, so since the directrix is at y = 2,then the part of the hyperbola that is closest to this focus will open downwards and the equation is given by;

r = ed/(1 + e•sinθ)

Plugging in the relevant values, we have;

r = (3.5 × 2)/(1 + 3.5(sinθ))

r = 7/(1 + 3.5(sinθ))

To make every figure a whole number, let's multiply numerator and denominator by 2 to give;

r = 14/(2 + 7(sinθ))

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Can somebody please help me

Vinil7 [7]

Answer:

the answer to #9 is (½, 1½)

Step-by-step explanation:

to find the midpoint, you need to use the midpoint formula

the midpoint formula is (x₁ + x₂ / 2 , y₁ + y₂ / 2)

for #9 you have to plug in the appropriate numbers in the formula

(-4 + 5 / 2 , 4 - 1 / 2)

(½ , 1½)

3 0

3 years ago

Use the augmented matrix method to solve the following system of equations. Your answers may be given as decimals or fractions.

Alexandra [31]

Answer:

$x\ =\ \dfrac{21}{5}$

$y\ =\ \dfrac{3}{5}$

z = 2

Step-by-step explanation:

Given equations are

x - 2y - z = 2

x + 3y - 2z = 4

-x + 2y + 3z = 2

from the given equations the augmented matrix can be written as

$\left[\begin{array}{ccc}1&-2&-1:2\\1&3&-2:4\\-1&2&3:2\end{array}\right]$

$R_2=>R_2-R_1\ and\ R_3=>R_3+R_1$

$=\ \left[\begin{array}{ccc}1&-2&-1:2\\0&5&-1:2\\0&0&2:4\end{array}\right]$

$R_2=>\dfrac{R_2}{5}$

$=\ \left[\begin{array}{ccc}1&-2&-1:2\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\0&0&2:4\end{array}\right]$

$R_1=>R_1+2.R_2$

$=\ \left[\begin{array}{ccc}1&0&-1-\dfrac{2}{5}:2+\dfrac{4}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&2:4\end{array}\right]$

$=\ \left[\begin{array}{ccc}1&0&\dfrac{-7}{5}:\dfrac{14}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&2:4\end{array}\right]$

$R_3=>\dfrac{R_3}{2}$

$=\ \left[\begin{array}{ccc}1&0&\dfrac{-7}{5}:\dfrac{14}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&1:2\end{array}\right]$

$R_1=>R_1+\dfrac{7}{5}R_3\ and\ R_2+\dfrac{1}{5}R_3$

$=\ \left[\begin{array}{ccc}1&0&0:\dfrac{14}{5}+\dfrac{7}{5}\\\\0&1&0:\dfrac{2}{5}+\dfrac{1}{5}\\\\0&0&1:2\end{array}\right]$

So, from the above augmented matrix, we can write

$x\ =\ \dfrac{21}{5}$

$y\ =\ \dfrac{3}{5}$

z = 2

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3 years ago

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Answer:

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2 years ago

Peter buys a camper van costing £35000 + 20% VAT. He pays a 50% deposit and then arranges to pay the remaining balance in 20 equ

shutvik [7]

The amount paid each month is 1050.

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Next, divide 42,000 by 20, and you would get 1050.

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3 years ago

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