You just multiply them normally because there are only two numbers
(1) This series consists of terms of an arithmetic sequence:
179 - 173 = 6
173 - 167 = 6
and so on, so that the <em>n</em>-th term in the series is (for <em>n</em> ≥ 1)
<em>a(n)</em> = 179 - 6 (<em>n</em> - 1) = 185 - 6<em>n</em>
Then the sum of the first 53 terms is
![\displaystyle\sum_{n=1}^{53}(185-6n) = 185\sum_{n=1}^{53}1-6\sum_{n=1}^{53}n](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B53%7D%28185-6n%29%20%3D%20185%5Csum_%7Bn%3D1%7D%5E%7B53%7D1-6%5Csum_%7Bn%3D1%7D%5E%7B53%7Dn)
![\displaystyle\sum_{n=1}^{53}(185-6n) = 185\times53-6\times\frac{53\times54}2](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B53%7D%28185-6n%29%20%3D%20185%5Ctimes53-6%5Ctimes%5Cfrac%7B53%5Ctimes54%7D2)
![\displaystyle\sum_{n=1}^{53}(185-6n) = \boxed{1219}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B53%7D%28185-6n%29%20%3D%20%5Cboxed%7B1219%7D)
(2) This series has terms from a geometric sequence:
-12 / 6 = -2
24/(-12) = -2
-48/24 = -2
and so on. The <em>n</em>-th term is (again, for <em>n</em> ≥ 1)
<em>a(n)</em> = 6 (-2)ⁿ⁻¹
and the sum of the first 19 terms is
![\displaystyle\sum_{n=1}^{19}6(-2)^{n-1} = 6\left(1 + (-2) + (-2)^2 + (-2)^3 + \cdots+(-2)^{19}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B19%7D6%28-2%29%5E%7Bn-1%7D%20%3D%206%5Cleft%281%20%2B%20%28-2%29%20%2B%20%28-2%29%5E2%20%2B%20%28-2%29%5E3%20%2B%20%5Ccdots%2B%28-2%29%5E%7B19%7D%5Cright%29)
Multiply both sides by -2 :
![\displaystyle-2\sum_{n=1}^{19}6(-2)^{n-1} = 6\left((-2) + (-2)^2 + (-2)^3 + (-2)^4 + \cdots+(-2)^{20}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle-2%5Csum_%7Bn%3D1%7D%5E%7B19%7D6%28-2%29%5E%7Bn-1%7D%20%3D%206%5Cleft%28%28-2%29%20%2B%20%28-2%29%5E2%20%2B%20%28-2%29%5E3%20%2B%20%28-2%29%5E4%20%2B%20%5Ccdots%2B%28-2%29%5E%7B20%7D%5Cright%29)
Subtracting this from the first sum gives
![\displaystyle(1-(-2))\sum_{n=1}^{19}6(-2)^{n-1} = 6\left(1 -(-2)^{20}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%281-%28-2%29%29%5Csum_%7Bn%3D1%7D%5E%7B19%7D6%28-2%29%5E%7Bn-1%7D%20%3D%206%5Cleft%281%20-%28-2%29%5E%7B20%7D%5Cright%29)
and solving for the sum, you get
![\displaystyle3\sum_{n=1}^{19}6(-2)^{n-1} = 6\left(1 -(-2)^{20}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle3%5Csum_%7Bn%3D1%7D%5E%7B19%7D6%28-2%29%5E%7Bn-1%7D%20%3D%206%5Cleft%281%20-%28-2%29%5E%7B20%7D%5Cright%29)
![\displaystyle\sum_{n=1}^{19}6(-2)^{n-1} = 2\left(1 -(-2)^{20}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B19%7D6%28-2%29%5E%7Bn-1%7D%20%3D%202%5Cleft%281%20-%28-2%29%5E%7B20%7D%5Cright%29)
![\displaystyle\sum_{n=1}^{19}6(-2)^{n-1} = 2\left(1 -(-1)^{20}2^{20}\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B19%7D6%28-2%29%5E%7Bn-1%7D%20%3D%202%5Cleft%281%20-%28-1%29%5E%7B20%7D2%5E%7B20%7D%5Cright%29)
![\displaystyle\sum_{n=1}^{19}6(-2)^{n-1} = 2\left(1 -2^{20}\right) = 2-2^{21} = \boxed{-2,097,150}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bn%3D1%7D%5E%7B19%7D6%28-2%29%5E%7Bn-1%7D%20%3D%202%5Cleft%281%20-2%5E%7B20%7D%5Cright%29%20%3D%202-2%5E%7B21%7D%20%3D%20%5Cboxed%7B-2%2C097%2C150%7D)
You can do this by replacing x = -2y + 10 in second equation and solving for y then plug this value of y in first equation to find x then use these values of x and y in last equation to find zSHOW FULL ANSWER
Answer:
Step-by-step explanation:
∠ADE = ∠CAB -34
∠ADE = (1/) of ∠CAB because of alternate interior angles of parallel lines
(1/2)∠CAB = ∠CAB - 34 => ∠CAB = 68° and ∠ADE = 34°
A)3 : 2
1080 : x
1 part = 360
2 *360 = 720
b) 3 :2
x:252
1 part = 126
126*3 = 378
:)