HELPPPP PLEASE IN NEED HELP FAST I NEED ALL 5 DONE

For what values of θ on the polar curve r=θ, with 0≤θ≤2π , are the tangent lines horizontal? Vertical?

Bond [772]

Given that $r=\theta$ , then $r'=1$

The slope of a tangent line in the polar coordinate is given by:

$m= \frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}$

Thus, we have:

$m= \frac{\sin\theta+\theta\cos\theta}{\cos\theta-\theta\sin\theta}$

Part A:

For horizontal tangent lines, m = 0.

Thus, we have:

$\sin\theta+\theta\cos\theta=0 \\ \\ \theta\cos\theta=-\sin\theta \\ \\ \theta=- \frac{\sin\theta}{\cos\theta} =-\tan\theta$

Therefore, the values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:

θ = 0

θ = 2.02875783811043

θ = 4.91318043943488

Part B:

For vertical tangent lines, $\frac{1}{m} =0$

Thus, we have:

$\cos\theta-\theta\sin\theta=0 \\ \\ \Rightarrow\theta\sin\theta=\cos\theta \\ \\ \Rightarrow\theta= \frac{\cos\theta}{\sin\theta} =\sec\theta$

Therefore, the values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:

θ = 4.91718592528713

3 0

3 years ago

The graph below represents the solution set of which inequality?

natulia [17]

Answer:

option: B ( $x^2+2x-8$ ) is correct.

Step-by-step explanation:

We are given the solution set as seen from the graph as:

(-4,2)

1)

On solving the first inequality we have:

$x^2-2x-8$

On using the method of splitting the middle term we have:

$x^2-4x+2x-8$

⇒ $x(x-4)+2(x-4)=0$

⇒ $(x+2)(x-4)$

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

$x+2>0$ and $x-4$

i.e. x>-2 and x<4

so we have the region as:

(-2,4)

Case 2:

$x+2$ and $x-4>0$

i.e. x<-2 and x>4

Hence, we did not get a common region.

Hence from both the cases we did not get the required region.

Hence, option 1 is incorrect.

2)

We are given the second inequality as:

$x^2+2x-8$

On using the method of splitting the middle term we have:

$x^2+4x-2x-8$

⇒ $x(x+4)-2(x+4)$

⇒ $(x-2)(x+4)$

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

$x-2>0$ and $x+4$

i.e. x>2 and x<-4

Hence, we do not get a common region.

Case 2:

$x-2$ and $x+4>0$

i.e. x<2 and x>-4

Hence the common region is (-4,2) which is same as the given option.

Hence, option B is correct.

3)

$x^2-2x-8>0$

On using the method of splitting the middle term we have:

$x^2-4x+2x-8>0$

⇒ $x(x-4)+2(x-4)>0$

⇒ $(x-4)(x+2)>0$

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

$x+2>0$ and $x-4>0$

i.e. x>-2 and x>4

Hence, the common region is (4,∞)

Case 2:

$x+2$ and $x-4$

i.e. x<-2 and x<4

Hence, the common region is: (-∞,-2)

Hence, from both the cases we did not get the desired answer.

Hence, option C is incorrect.

4)

$x^2+2x-8>0$

On using the method of splitting the middle term we have:

$x^2+4x-2x-8>0$

⇒ $x(x+4)-2(x+4)>0$

⇒ $(x-2)(X+4)>0$

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

$x-2$ and $x+4$

i.e. x<2 and x<-4

Hence, the common region is: (-∞,-4)

Case 2:

$x-2>0$ and $x+4>0$

i.e. x>2 and x>-4.

Hence, the common region is: (2,∞)

Hence from both the case we do not have the desired region.

Hence, option D is incorrect.

5 0

3 years ago

Find the equation of a line with m = 3 that passes though the point (-12, 4).

Viefleur [7K]

Answer:

y = 3x + 40

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = 3 , then

y = 3x + c ← is the partial equation

To find c substitute (- 12, 4 ) into the partial equation

4 = - 36 + c ⇒ c = 4 + 36 = 40

y = 3x + 40 ← equation of line

8 0

3 years ago

The diagonal of the floor of a square room is 6 meters.what is the length of a side of the room? round to the nearest tenth.

shutvik [7]

Side of the room = x.
From the right triangle, where sides of the square are legs of the right triangle, and the diagonal of the square is a hypotenuse of the right triangle.
x²+x²=6²
2x²=36
x²=18
x=√18 =√(2*9)=3√2≈4.2m

4 0

3 years ago

The midpoint of a circle is (3,2), the

Marysya12 [62]

yo sup??

mid point of circle=centre of circle

let centre be O(3,2)

and the point be P(-4,8) and the other point be Q(x,y)

by mid point formula

-4+x/2=3 and 8+y/2=2

x=10 and y=-4

therefore coordinates of the other point are (10,-4)

Hope this helps.

3 0

3 years ago