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3 years ago

Convert the equation r = 12 sin theta to Cartesian coordinates. Describe the resulting curve.

Mathematics

1 answer:

VLD [36.1K]3 years ago

7 0

Answer:

Step-by-step explanation:

Given that there is a polar equation as

$r=12 sin \theta$

This has to be converted into cartesian.

We know the conversion is

$r^2 =x^2+y^2 \\tan \theta = \frac{y}{x}$

Using this we can say that

$sin^2 \theta = 1-cos^2 \theta \\= 1-\frac{1}{sec^2 \theta} \\=1-\frac{1}{1+tan^2 \theta} \\=1-\frac{1}{1+\frac{y^2}{x^2} } \\=1-\frac{x^2}{x^2+y^2} \\=\frac{y^2}{y^2+y^2}$

$\sqrt{x^2+y^2} =12(\frac{y}{\sqrt{x^2+y^2} } \\x^2+y^2 =12y\\x^2+y^2-12y+36 = 36\\x^2+(y-6)^2 = 6^2$ }

Circle with centre (0,6) and radius 6.

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3 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

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3 years ago

The air force reports that the distribution of heights of male pilots is approximately normal, with a mean of 72.6 inches and a

Phantasy [73]

Answer:

Step-by-step explanation:

Given that X - the distribution of heights of male pilots is approximately normal, with a mean of 72.6 inches and a standard deviation of 2.7 inches.

Height of male pilot = 74.2 inches

We have to find the percentile

X = 74.2

Corresponding Z score = 74.2-72.6 = 1.6

P(X<174.2) = P(Z<1.6) = 0.5-0.4452=0.0548=5.48%

i.e. only 5% are below him in height.

Thus the malepilot is in 5th percentile.

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3 years ago