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bogdanovich [222]
3 years ago
8

Analyze the graph of the given polar curve. If possible, describe the shape of the graph (circle,rose curve, limacon, etc), and

state the domain, range, and maximum r-value of the graph. State whether the graph is continuous and whether is is bounded. Describe any symmetry the graph has. Give the equations of any asymptotes or state that the graph has no asymptotes. r= -2cos5theta
Mathematics
1 answer:
Darya [45]3 years ago
6 0

Rose petal with 5 petals

Domain: all real numbers

Range: [-2,2]

Symmetric about the x-axis

Continuous

Bounded

Maximum r-value: 2

No asymptotes

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Phil received a prize of xxx dollars from a poker tournament. The tournament cost him 100100100 dollars to enter.
nikitadnepr [17]

Answer:

$(x-100)

Step-by-step explanation:

  • Phil entered the poker tournament with $100.
  • He received a price of $x.

His net winnings from the tournament will be the price minus the entry fee.

Therefore:

Phil's net winnings=$(x-100)

8 0
3 years ago
In triangle ABC, what is m angle C
Degger [83]

Answer:

83.5 degrees

Step-by-step explanation:

first, find x:

x+5x-13+4x=180

10x=193

x=19.3

Then, substitute that value for c:

5*19.3-13

=83.5 degrees

4 0
2 years ago
Which is the standard form of the equation of the parabola that has a vertex of (3, 1) and a directric of x = -2?
vagabundo [1.1K]
Check the picture below.  So,the parabola looks like so, notice the distance "p". since the parabola is opening to the right, then "p" is positive, thus is 5.

\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
\boxed{(y-{{ k}})^2=4{{ p}}(x-{{ h}})} \\\\
(x-{{ h}})^2=4{{ p}}(y-{{ k}}) \\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\

\bf -------------------------------\\\\
\begin{cases}
h=3\\
k=1\\
p=5
\end{cases}\implies (y-1)^2=4(5)(x-3)\implies (y-1)^2=20(x-3)
\\\\\\
\cfrac{1}{20}(y-1)^2=x-3\implies \boxed{\cfrac{1}{20}(y-1)^2+3=x}

8 0
3 years ago
Read 2 more answers
If f(x) = 2x^4, then f'(3) =
valkas [14]

To find f'(3) (f prime of 3), you must find f' first.  f' is the derivative of the function f(x).

Finding the derivative of f(x) = 2x⁴ requires the use of the power rule.

The power rule for derivatives is \frac{d}{dx} [x^{n} ] =nx^{n-1}.  In other words, you bring the exponent forward and multiply it by the coefficient of the term, and then you subtract 1 from the original exponent.

f'(x) = \frac{d}{dx}(2x⁴)

f'(x) = 2(4)x³

f'(x) = 8x³

Now, to find f'(3), plug 3 into your derivative.

f'(3) = 8(3)³

f'(3) = 216

<h3>Answer:</h3>

f'(3) = 216

6 0
2 years ago
peanuts in bulk bins at the grocery store costs $0.30 per pound how much would 2 pounds of peanuts cost
Helen [10]
$0.60 cents would be the cost of 2 lbs
8 0
3 years ago
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