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Mamont248 [21]
3 years ago
12

Draw a graph of the rose curve. r = 4 cos 2θ, 0 ≤ θ ≤ 2π

Mathematics
2 answers:
satela [25.4K]3 years ago
6 0

Answer:

Attachment

Step-by-step explanation:

Given: r=4\cos2\theta

where, 0\leq \theta \leq 2\pi

We need to draw the graph of polar curve.

r=a\cos n\theta

It is flower shape graph.

whose number of leaf would be n

Here, we have polar equation

r=4\cos2\theta

Coefficient of Ф is 2

Number of leaf should be 4. We will draw the graph using graphing calculator.

Please see the attachment for graph.

alukav5142 [94]3 years ago
3 0
See the attachment below gor graph
hope that helps 

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Answer:

2.75 hours

Step-by-step explanation:

Speed is 24km per hour (\frac{24km}{1hour}), and the bus covered 66km.

We need to get a value of <em>km</em> with the rate (24/1) and distance (66) given.

We get:  \frac{1hour}{24km} *66km.  

The km cancel out, leaving us with 66/24 hours, or 2.75 hours.

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Hector can paint (128
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Step-by-step explanation:

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A particular frozen yogurt has 75 cal in 2 ounces how many calories are in 8 ounces of the yogurt
Vesnalui [34]
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3 years ago
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55. If 3x = 4y, the value of (x + y)^2 : (x - y)^2 is:<br>​
ladessa [460]

Answer:

\large\boxed{(x+y)^2:(x-y)^2=49}

Step-by-step explanation:

3x=4y\qquad\text{subtract}\ 3y\ \text{from both sides}\\\\3x-3y=y\qquad\text{distributive}\\\\3(x-y)=y\qquad\text{divide both sides by 3}\\\\x-y=\dfrac{y}{3}\qquad(*)\\------------------\\3x=4y\qquad\text{add}\ 3y\ \text{to both sides}\\\\3x+3y=7y\qquad\text{distributive}\\\\3(x+y)=7y\qquad\text{divide both sides by 3}\\\\x+y=\dfrac{7y}{3}\qquad(**)\\------------------

(x+y)^2:(x-y)^2=\dfrac{(x+y)^2}{(x-y)^2}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\=\left(\dfrac{x+y}{x-y}\right)^2\qquad\text{substitute}\ (*)\ \text{and}\ (**)\\\\=\left(\dfrac{\frac{7y}{3}}{\frac{y}{3}}\right)^2=\left(\dfrac{7y}{3}\cdot\dfrac{3}{y}\right)^2\qquad\text{cancel}\ 3\ \text{and}\ y\\\\=(7)^2=49

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3 years ago
Find the value of x. ​
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Sin x=24/29
Therefore x=sin^-1 24/29
5 0
3 years ago
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