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katrin2010 [14]
3 years ago
6

the angle theta is in the second quadrant and cos theta = -2/√29 determine possible coordinates for point P on the terminal arm

of theta a. (2,5) b. (-2,√29) c. (-5,2) d. (-2,5)
Mathematics
1 answer:
Gekata [30.6K]3 years ago
6 0

\cos(\theta)=-\frac{2}{\sqrt{29}} and $\theta$ lies in $2^{\text{th}}$ quadrant.

where, $x-$ coordinate is negative, and $y-$ coordinate is positive

so it can't a.

now, cosine means, side adjacent over the hypotenuse, in Cartesian plane, that will be $x-$ coordinate over the distance from origin.

Assume the triangle , with base $2$ units and hypotenuse $\sqrt{29}$ and it's in second quadrant. (so \cos(\theta)=-\frac{2}{\sqrt{29}})

now, the leftmost point on $x-$ axis is , obviously $(-2,0)$

and by Pythagoras theorem, we can find the perpendicular side, that will be $y^2=(\sqrt{29})^2-(2)^2\implies y=5$

so the coordinates of the upper vertex is $(-2,5)$, each point lying on this "ray" should have equal ratio of respective coordinates. i.e. $\frac25=\left|\frac xy\right| $

and it should lie on second quadrant, so $x<0 \, y>0$

Option d satisfies this.

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lisabon 2012 [21]

Answer:

After 16 rounds the champion of Ping-Pong player can be determined.

Step-by-step explanation:

This is an example of geometric progression,

First term = 65,536

\texttt{Common ratio = }\frac{1}{2}

Now we need to find which term is 1 in this GP.

N th term of GP is given by

             t_n=ar^{n-1}

Substituting

               1=65536\times \left ( \frac{1}{2}\right )^{n-1}\\\\\frac{1}{65536}=\left ( \frac{1}{2}\right )^{n-1}\\\\2^{n-1}=65536\\\\2^{n-1}=2^{16}\\\\n=17

Seventeenth term of this GP is 1.

That is after 16 rounds the champion of Ping-Pong player can be determined.

3 0
3 years ago
Find the missing dimension of each prism.
Leto [7]

Given:

Volume of the prism = 60 in³

Breadth = 2.5 in

Height = 4 in.

To find:

The length of the prism.

Solution:

We know that, volume of a prism is

V=l\times b\times h

where, l is length, b is breadth and h is height.

Putting V=60, b=2.5 and h=4, we get

60=l\times 2.5\times 4

60=l\times 10

Divide both sides by 10.

\dfrac{60}{10}=l

6=l

Therefore, the missing dimension of the prism is 6 in.

7 0
3 years ago
Find the domain and range of radical function
Tju [1.3M]

The radical function is 3√x + 1.

Since the cubic root of zero is zero, 0 would be x.

Add zero and one, which gives you 1. 1 is y.

You get (0,1)

Since this is a positive radical, it would be going to the top right.

So, the domain is x ≥ 0

the range is y ≥ 1

8 0
3 years ago
A movie theater has a 24​-foot-high screen located 6 feet above your eye level. If you sit x feet back from the​ screen, your vi
pantera1 [17]

Answer:

The viewing angles are as follows:

For x=5 feet, θ = 0.529 radians

For x=10 feet, θ = 0.708 radians

For x=15 feet, θ = 0.726 radians

For x=20 feet, θ = 0.691 radians

For x=25 feet, θ = 0.640 radians

Step-by-step explanation:

The viewing angle is given as:

θ = tan⁻¹(30/x) - tan⁻¹ (6/x)

where x is the distance between you and the screen.

The question is asking us to find the viewing angle θ at various distances. The distance value needs to be substituted in the above equation in place of x. So,

<u>For x=5 feet:</u>

θ = tan⁻¹(30/5) - tan⁻¹ (6/5)

  = 1.4056 - 0.8761

θ = 0.529 radians

<u>For x = 10 feet:</u>

θ = tan⁻¹(30/10) - tan⁻¹ (6/10)

  = 1.249 - 0.540

θ = 0.708 radians

<u>For x = 15 feet:</u>

θ = tan⁻¹(30/15) - tan⁻¹ (6/15)

  = 1.107 - 0.380

θ = 0.726 radians

<u>For x = 20 feet:</u>

θ = tan⁻¹(30/20) - tan⁻¹ (6/20)

  = 0.983 - 0.291

θ = 0.691 radians

<u>For x = 25 feet:</u>

θ = tan⁻¹(30/25) - tan⁻¹ (6/25)

  = 0.876 - 0.235

θ = 0.640 radians

5 0
3 years ago
Select Yes or No to state whether each data set is likely to be normally distributed.
kenny6666 [7]

we know that

Normal distribution:

The normal distribution is a very common continuous probability distribution

A probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean

now, we will check each options

option-A:

we know that number of coupons will always be counted

so, it is discrete in nature

so, this is not normal distribution

option-B:

we know that

the weights of the pumpkins that are delivered to a supermarket can be any values

so, this will be continuous in nature

so, this is normal distribution

option-C:

the number of raisins in each 8-oz box of raisins at a supermarket

number will always be counted

so, this is discrete in nature

so, this is not normal distribution

option-D:

the amount of time customers spend waiting in the checkout line at a supermarket

time can be decimal or number

so, this is continuous in nature

so, this is normal distribution

5 0
3 years ago
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