The coordinates of a point is the location of the point in a plane.
<em>The coordinates of the centers of holes are: (48.5, 28) and (28, 48.5)</em>
Given
![\theta_1 = \theta_2 = \theta_3 = 30^o](https://tex.z-dn.net/?f=%5Ctheta_1%20%3D%20%5Ctheta_2%20%3D%20%5Ctheta_3%20%3D%2030%5Eo)
![R = 60](https://tex.z-dn.net/?f=R%20%3D%2060)
![r = 56](https://tex.z-dn.net/?f=r%20%3D%2056)
I've added an attachment as an illustration
<u />
<u>Considering </u>
<u />
To solve for x1, we make use of cosine ratio.
So, we have:
![\cos(\theta_1) =\frac{x}{r}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta_1%29%20%3D%5Cfrac%7Bx%7D%7Br%7D)
Make x the subject
![x_1 = r \times \cos(\theta_1)](https://tex.z-dn.net/?f=x_1%20%3D%20r%20%5Ctimes%20%5Ccos%28%5Ctheta_1%29)
![x_1 = 56 \times \cos(30^o)](https://tex.z-dn.net/?f=x_1%20%3D%2056%20%5Ctimes%20%5Ccos%2830%5Eo%29)
![x_1 = 48.5](https://tex.z-dn.net/?f=x_1%20%3D%2048.5)
To solve for y1, we make use of sine ratio.
So, we have:
![\sin(\theta_1) =\frac{y_1}{r}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta_1%29%20%3D%5Cfrac%7By_1%7D%7Br%7D)
Make y the subject
![y_1 = r \times \sin(\theta_1)](https://tex.z-dn.net/?f=y_1%20%3D%20r%20%5Ctimes%20%5Csin%28%5Ctheta_1%29)
![y_1 = 56 \times \sin(30^o)](https://tex.z-dn.net/?f=y_1%20%3D%2056%20%5Ctimes%20%5Csin%2830%5Eo%29)
![y_1 = 28](https://tex.z-dn.net/?f=y_1%20%3D%2028)
So, we have:
![(x_1,y_1) = (48.5,28)](https://tex.z-dn.net/?f=%28x_1%2Cy_1%29%20%3D%20%2848.5%2C28%29)
<u>Considering </u>
<u />
To solve for x2, we make use of cosine ratio.
So, we have:
![\cos(\theta_1+\theta_2) =\frac{x_2}{r}](https://tex.z-dn.net/?f=%5Ccos%28%5Ctheta_1%2B%5Ctheta_2%29%20%3D%5Cfrac%7Bx_2%7D%7Br%7D)
Make x the subject
![x_2 = r \times \cos(\theta_1+\theta_2)](https://tex.z-dn.net/?f=x_2%20%3D%20r%20%5Ctimes%20%5Ccos%28%5Ctheta_1%2B%5Ctheta_2%29)
![x_2 = 56 \times \cos(30+30)](https://tex.z-dn.net/?f=x_2%20%3D%2056%20%5Ctimes%20%5Ccos%2830%2B30%29)
![x_2 = 56 \times \cos(60^o)](https://tex.z-dn.net/?f=x_2%20%3D%2056%20%5Ctimes%20%5Ccos%2860%5Eo%29)
![x_2 = 28](https://tex.z-dn.net/?f=x_2%20%3D%2028)
To solve for y1, we make use of sine ratio.
So, we have:
![\sin(\theta_1+\theta_2) =\frac{y_2}{r}](https://tex.z-dn.net/?f=%5Csin%28%5Ctheta_1%2B%5Ctheta_2%29%20%3D%5Cfrac%7By_2%7D%7Br%7D)
Make y the subject
![y_2 = r \times \sin(\theta_1+\theta_2)](https://tex.z-dn.net/?f=y_2%20%3D%20r%20%5Ctimes%20%5Csin%28%5Ctheta_1%2B%5Ctheta_2%29)
![y_2 = 56 \times \sin(30+30)](https://tex.z-dn.net/?f=y_2%20%3D%2056%20%5Ctimes%20%5Csin%2830%2B30%29)
![y_2 = 56 \times \sin(60)](https://tex.z-dn.net/?f=y_2%20%3D%2056%20%5Ctimes%20%5Csin%2860%29)
![y_2 = 48.5](https://tex.z-dn.net/?f=y_2%20%3D%2048.5)
So, we have:
![(x_2,y_2) = (28,48.5)](https://tex.z-dn.net/?f=%28x_2%2Cy_2%29%20%3D%20%2828%2C48.5%29)
<em>Hence, the coordinates of the centers of the holes are: (48.5, 28) and (28, 48.5)</em>
Read more about coordinate geometry at:
brainly.com/question/8121530
They do not go through the origin. (Answer Choice C)
I believe your answer is 48. I hope i was helpful
Answer:
154,897 REGISTERED LABRADOR RETRIEVERS WERE THERE
Step-by-step explanation:
Given the statement: IN 1999, THERE WERE 9860 GREAT DANES REGISTERED WITH THE AMERICAN KENNEL CLUB.
Also, THE NUMBER OF REGISTERED LABRADOR RETRIEVERS WAS 6997 MORE THAN FIFTEEN TIMES THE NUMBER OF REGISTERED GREAT DANES
Let x represents the number of Great Danes registered and y represents the number of Labrador retrievers registered.
The statement:
"Fifteen times the number of registered great Danes " means 15x
and
"6997 more than fifteen times the number of registered great Danes" means 15x + 6997
then as per the given statement:
x = 9860 .
To solve for y;
y = 15x+ 6997
Substitute the value of x we get;
Simplify:
y = 154,897
Therefore, the registered Labrador Retrievers were there are; 154,897
douwdek0 and 16 more users found this answer helpful
To find 1/6 of the time, you will need to convert the hours into minutes. So 3x60 minutes equals 180 minutes. Now that you have that converted into minutes, you can divide the 180 by 6.
180/6 = 30. So Cody has watched 30 minutes of the movie.
30 minutes is 1/2 of an hour.