Easy 25 points answer soon please

Find an equation in rectangular coordinates for the cylindrical equation r = 2sintheta

frutty [35]

Answer:

$x^2+(y-1)^2=1$

Step-by-step explanation:

The given cylindrical equation is

$r=2\sin \theta$

Multiply both sides by r.

$r^2=2r\sin \theta$ .... (1)

The required formulas are

$x=r\cos \theta,y=r\sin \theta,r^2=x^2+y^2$

Substitute $r\sin \theta =y,r^2=x^2+y^2$ in equation (1).

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

$x^2+y^2-2y=0$

Add 1 on both sides.

$x^2+(y^2-2y+1)=1$

$(x-0)^2+(y-1)^2=1^2$ $[\because (a-b)^2=a^2-2ab+b^2]$

It is the equation of a circle centered at (0,1) with radius 1.

Therefore, the equation in rectangular coordinates is $x^2+(y-1)^2=1$ .

3 0

3 years ago

If 7 is added to a number then it becomes at least 15 what is the number?

Whitepunk [10]

You’re answer will be 8 ;)

8 0

3 years ago

Read 2 more answers

For a certain candy, 20% of the pieces are yellow, 15% are red, 20% are blue, 20% are green, and the rest are brown. a

OLEGan [10]

Answer:

Step-by-step explanation:

Based on the question we are given the percentages of each of the types of candies in the bag except for brown. Since the sum of all the percentages equals 75% and brown is the remaining percent then we can calculate that brown is (100-75 = 25%) 25% of the bag. Now we can show the probabilities of getting a certain type of candy by placing the percentages over the total percentage (100%).

Brown: $\frac{25}{100}$
Yellow or Blue: $\frac{20}{100} +\frac{20}{100} = \frac{40}{100}$ ....add the numerators

Not Green: $\frac{80}{100}$ .... since the sum of all the rest is 80%

Stiped: $\frac{25}{100}$ .... there are 0 striped candies.

Assuming the <u><em>ratios/percentages</em></u> of the candies stay the same having an infinite amount of candy will not affect the probabilities. That being said in order to calculate consecutive probability of getting 3 of a certain type in a row we have to multiply the probabilities together. This is calculated by multiplying the numerators with numerators and denominators with denominators.