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2 years ago

Find the polar equation of the conic with the focus at the pole, directrix x = -5, and eccentricity 2.

Mathematics

1 answer:

prohojiy [21]2 years ago

3 0

Step-by-step explanation:

Given that,

Directrix, d= -5, and eccentricity, e = 2.

The polar equation of the conic is given by :

$r=\dfrac{ed}{1-e\cos\theta}$

Put all the values,

$r=\dfrac{-5\times 2}{1-2\cos\theta}\\\\=\dfrac{-10}{1-2\cos\theta}$

As e>1, it is a hyperbola. Hence, this is the required solution.

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A store owner paid $120 wholesale price for a suit. He needs to make a 35% profit on

Juliette [100K]

Answer:

$162 to make a 35% profit

Step-by-step explanation:

35% of 120 is 42. Add the 42 to 120.

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2 years ago

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If A is the area of a circle with radius r and the circle expands as time passes, find dA/dt in terms of dx/dt

valina [46]

The area of a circle in terms of the circle's radius $r$ is

$A=\pi r^2$

Differentiating both sides with respect to $t$ , you get

$\dfrac{\mathrm dA}{\mathrm dt}=2\pi r\dfrac{\mathrm dr}{\mathrm dt}$

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3 years ago

hi, i dont undertand number 20 because i was absent in class today and i rerally need help, i will appraciate with the help, and

Mariulka [41]

Given:

The equation is,

$2\log _3x-\log _3(x-2)=2$

Explanation:

Simplify the equation by using logarthimic property.

$\begin{gathered} 2\log _3x-\log _3(x-2)=2 \\ \log _3x^2-\log _3(x-2)=2_{}\text{ \lbrack{}log(a)-log(b) = log(a/b)\rbrack} \\ \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \end{gathered}$

Simplify further.

$\begin{gathered} \log _3\lbrack\frac{x^2}{x-2}\rbrack=2 \\ \frac{x^2}{x-2}=3^2 \\ x^2=9(x-2) \\ x^2-9x+18=0 \end{gathered}$

Solve the quadratic equation for x.

$\begin{gathered} x^2-6x-3x+18=0 \\ x(x-6)-3(x-6)=0 \\ (x-6)(x-3)=0 \end{gathered}$

From the above equation (x - 6) = 0 or (x - 3) = 0.

For (x - 6) = 0,

$\begin{gathered} x-6=0 \\ x=6 \end{gathered}$

For (x - 3) = 0,

$\begin{gathered} x-3=0 \\ x=3 \end{gathered}$

The values of x from solving the equations are x = 3 and x = 6.

Substitute the values of x in the equation to check answers are valid or not.

For x = 3,

$\begin{gathered} 2\log _3(3^{})-\log _3(3-2)=2 \\ 2\log _33-\log _31=2 \\ 2\cdot1-0=2 \\ 2=2 \end{gathered}$

Equation satisfy for x = 3. So x = 3 is valid value of x.

For x = 6,

$\begin{gathered} 2\log _36-\log _3(6-2)=2 \\ 2\log _36-\log _34=2 \\ \log _3(6^2)-\log _34=2 \\ \log _3(\frac{36}{4})=2 \\ \log _39=2 \\ \log _3(3^2)=2 \\ 2\log _33=2 \\ 2=2 \end{gathered}$

Equation satifies for x = 6.

Thus values of x for equation are x = 3 and x = 6.

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1 year ago

A tailor pays $10.25 per yard for wool and pays $5.75 per yard for cotton. He bought 4.8 yards of cotton and spent a total of $8

Alex777 [14]

Answer:

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Step-by-step explanation:

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3 years ago

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Aleksandr [31]

Answer:

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Step-by-step explanation:

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2 years ago