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3 years ago

Determine the zeros of r=2sin5theta

Mathematics

1 answer:

vazorg [7]3 years ago

6 0

Answer:

$\theta=\frac{n\pi}{5}$

Step-by-step explanation:

You have the following function:

$r=2sin5\theta$ (1)

In order to find the zeros of the function you equal to zero the equation (1), and then you solve for θ:

$2sin5\theta=0\\\\sin5\theta=0\\\\5\theta=sin^{-1}(0)=n\pi;\ \ \ \ n=0,1,2,3,..\\\\\theta=\frac{n\pi}{5}$

Then, there are infinite zeros for the function of the equation (1), because n has infinite positive integers values.

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The engineers designing the All Aboard Railroad between Boca Raton and Jupiter decide to create parallel tracks through this por

olga2289 [7]

Answer:

there are no signs between the x and y and constant

it could be

2x+5y=15

2x+5y=-15

-2x+5y=15

2x-5y=15

for ax+by=c, the equation of a line paralell to that is

ax+by=d where a=a, b=b, and c and d are constants

(for this answer, I'm going to use 2x+5y=15)

given 2x+5y=15, the equation of a line paralell to that is 2x+5y=d

to find d, subsitute the point (4,-2), basically put 4 in for x and -2 for y to get the constant

2x+5y=d

2(4)+5(-2)=d

8-10=d

-2=d

the eqaution is 2x+5y=-2 (Only if the original equation is 2x+5y=-15

pls mark me brainlest

5 0

3 years ago

H + 169 = 759 solve for h

steposvetlana [31]

Answer:

h = 590

Step-by-step explanation:

To find x, rearrange the equation to where h is on its own side.

Rearranged equation:

$759 - 169 = h$

Since 169 added to h was to find 769, 169 subtracted from 759 represents h.

Solve:

$759 - 169 = 590$

590 = h.

5 0

3 years ago

What is the value of x? Enter your answer in the box. x =

Semmy [17]

Answer:

Step-by-step explanation:

8 0

3 years ago

Each hour, the number of bacteria in Dr. Nall’s petri dish tripled. What percent is the population compared to the population th

dimulka [17.4K]

THE ANSWER IS THE D!!!!!!!!!!!!

3 0

4 years ago

Read 2 more answers

Find all possible values of α+

const2013 [10]

Answer:

$\rm\displaystyle 0,\pm\pi$

Step-by-step explanation:

please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation

===========================

we want to find all possible values of α+β+γ when tanα+tanβ+tanγ = tanαtanβtanγ to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma )$

factor out tanγ:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)$

divide both sides by tanαtanβ-1 and that yields:

$\rm\displaystyle \tan( \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) - 1}$

multiply both numerator and denominator by-1 which yields:

$\rm\displaystyle \tan( \gamma ) = - \bigg(\frac{ \tan( \alpha ) + \tan( \beta ) }{ 1 - \tan( \alpha ) \tan( \beta ) } \bigg)$

recall angle sum indentity of tan:

$\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta )$

let α+β be t and transform:

$\rm\displaystyle \tan( \gamma ) = - \tan( t)$

remember that tan(t)=tan(t±kπ) so

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi )$

therefore when k is 1 we obtain:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi )$

remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi )$

recall that if we have common trigonometric function in both sides then the angle must equal which yields:

$\rm\displaystyle \gamma = - \alpha - \beta \pm \pi$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }$

when is 0:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 )$

likewise by Opposite Angle Identity we obtain:

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 )$

recall that if we have common trigonometric function in both sides then the angle must equal therefore:

$\rm\displaystyle \gamma = - \alpha - \beta \pm 0$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ 0 }$

and we're done!

8 0

3 years ago

Read 2 more answers