Answer:
Disclaimer: There is error in the Q. In (II) there should have been 2x - 3y = 12
5x + 3y = 9 -----(I)
2x - 3y = 12 ----- (II)
Add (I) and (II)
7x = 21
x = 3
Putting the value of x = 3 in (I) we get
5(3)+3y=9⇒15+3y=9⇒3y=9-15=-6⇒y=-2
Thus, (x, y) = (3, -6).
Step-by-step explanation:
Answer:
Can you please include the complete question?
Step-by-step explanation:
Answer:
The answer is option C.
<h3>-6, -5 2/5, -4 1/5</h3>
Step-by-step explanation:
The arithmetic sequence is given by
![A(n) = - 6 + (n - 1)( \frac{1}{5} )](https://tex.z-dn.net/?f=A%28n%29%20%3D%20%20-%206%20%2B%20%28n%20-%201%29%28%20%5Cfrac%7B1%7D%7B5%7D%20%29)
where n is the number of terms
<u>For</u><u> </u><u>the</u><u> first</u><u> </u><u>term</u>
n = 1
So we have
![A(1) = - 6 + (1 - 1)( \frac{1}{5} )](https://tex.z-dn.net/?f=A%281%29%20%3D%20%20-%206%20%2B%20%281%20-%201%29%28%20%5Cfrac%7B1%7D%7B5%7D%20%29)
![= - 6 + (0)( \frac{1}{5} )](https://tex.z-dn.net/?f=%20%3D%20%20-%206%20%2B%20%280%29%28%20%5Cfrac%7B1%7D%7B5%7D%20%29)
![= - 6](https://tex.z-dn.net/?f=%20%3D%20%20-%206)
<u>For</u><u> </u><u>the</u><u> </u><u>fou</u><u>rth</u><u> term</u>
n = 4
![A(4) = - 6 + (4 - 1)( \frac{1}{5} )](https://tex.z-dn.net/?f=A%284%29%20%3D%20%20-%206%20%2B%20%284%20-%201%29%28%20%5Cfrac%7B1%7D%7B5%7D%20%29)
![= - 6 + (3)( \frac{1}{5} )](https://tex.z-dn.net/?f=%20%3D%20%20-%206%20%2B%20%283%29%28%20%5Cfrac%7B1%7D%7B5%7D%20%29)
![= - 5 \frac{2}{5}](https://tex.z-dn.net/?f=%20%3D%20%20-%205%20%5Cfrac%7B2%7D%7B5%7D%20)
<u>For</u><u> </u><u>the</u><u> </u><u>tenth</u><u> </u><u>term</u>
n = 10
![A(10) = - 6 + (10 - 1)( \frac{1}{5} )](https://tex.z-dn.net/?f=A%2810%29%20%3D%20%20-%206%20%2B%20%2810%20-%201%29%28%20%5Cfrac%7B1%7D%7B5%7D%20%29)
![= - 6 + (9)( \frac{1}{5} )](https://tex.z-dn.net/?f=%20%3D%20%20-%206%20%2B%20%289%29%28%20%5Cfrac%7B1%7D%7B5%7D%20%29)
![= - 4 \frac{1}{5}](https://tex.z-dn.net/?f=%20%3D%20%20-%204%20%5Cfrac%7B1%7D%7B5%7D%20)
Hope this helps you
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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