since it has a diameter of 28, then its radius must be half that or 14.
![\textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=14 \end{cases}\implies A=\pi (14)^2\implies A=196\pi ~\hfill \stackrel{\stackrel{semi-circle}{half~that}}{98\pi }](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D14%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%20%2814%29%5E2%5Cimplies%20A%3D196%5Cpi%20~%5Chfill%20%5Cstackrel%7B%5Cstackrel%7Bsemi-circle%7D%7Bhalf~that%7D%7D%7B98%5Cpi%20%7D)
Answer:
We use the formula:
S 10 = 8 · [(1/2)^9 - 1] / (1/2 - 1 );
Then, S 10 = 8 · ( 1/512 - 1 ) / ( 1/2 - 1 );
S 10 = 8 · ( - 511 / 512 ) / (-1 / 2 );
S 10 = ( - 511 / 64 ) · ( - 2 );
S 10 = + 511 / 32;
S 10 = 15.96;
Step-by-step explanation:
Step-by-step explanation:
3 - ( - 5 + 4)
4
- 1 - ( 0 - 2)
1
- 2 - ( 0 - 2)
0
Let
x-------------> <span>the price of each cup
</span>y-------------> the price of each plate
we know that
5x+3y=28.50----------> equation 1
and
x=y-6.30--------------> equation 2
I substitute 2 in 1
5*(y-6.30)+3y=28.50---------> 5y-31.5+3y=28.5-----> 8y=60
y=7.5
x=y-6.3---------> x=7.5-6.3---------> x=1.2
the answer is
the price of each cup is $1.2
the price of each plate is $7.5
