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The polar coordinates of a point are 3π 4 and 7.00 m. What are its Cartesian coordinates (in m)? (x, y) = 6.06,−3.5 m

Keith_Richards [23]

Answer:

a) $\left(x,y\right)=\left(4.95,-4.95\right)$

b) $r\angle\theta = 7\angle0.5236\,\text{radians}$

Step-by-step explanation:

Polar coordinates are represented as: $r\angle\theta$ , where 'r' is the length (or magnitude) of the line, and ' $\theta$ ' is the angle measured from the positive x-axis.

in our case:

$7\angle\dfrac{3\pi}{4}$

to covert the polar to cartesian:

$x = r\cos{\theta}$

$y = r\sin{\theta}$

we can plug in our values:

$x = 7\cos{\dfrac{3\pi}{4}} = -7\dfrac{\sqrt{2}}{2}$

$y = 7\sin{\dfrac{3\pi}{4}} = 7\dfrac{\sqrt{2}}{2}$

the Cartesian coordinates are:

$\left(x,y\right)=\left(-7\dfrac{\sqrt{2}}{2},7\dfrac{\sqrt{2}}{2}\right)$

$\left(x,y\right)=\left(4.95,-4.95\right)$

(b) to convert (x,y) = (6.06,-3.5)

we'll use the pythagoras theorem to find 'r'

$r^2 = x^2+y^2$

$r^2 = (6.06)^2+(-3.5)^2$

$r = \sqrt{48.97} \approx 7$

the angle can be found by:

$\tan{\theta} = \dfrac{y}{x}$

$\tan{\theta} = \dfrac{3.5}{6.06}$

$\theta = \arctan{left(\dfrac{3.5}{6.06}\right)}$

$\theta = 0.5236 \text{radians}$

to convert radians to degrees:

$\theta = 0.5236 \times \dfrac{180}{\pi} \approx 30^\circ$

writing in polar coordinates:

$r\angle\theta = 7\angle30^\circ\,\,\text{OR}\,\,7\angle0.5236\,\text{radians}$

5 0

2 years ago

The baker pays $0.80 per pound for sugar and $1.25 per pound for butter. Which of the following equations shows t, how much the

Lynna [10]

Answer:

(6 x $1.25) + (20 x $0.80)

5 0

2 years ago

Read 2 more answers

Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. 500

Alborosie

Answer:

The 95% confidence interval estimate for the true proportion of adults residents of this city who have cell phones is (0.81, 0.874).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.