first thing you have to do is find the rate for the bike so you divide 30 by 2 and you get 15 after you do that you make a problem to find the percentage so you take x/100 and 12/15 then you take 12 times 100 and that gives you 1200 then you divide by 15 and get 80. So the answer is 80%
Here you are. you can see my explication.
have fun
Hello there, multiply to each side of ratio until you get the answer, which in this case you have to multiply 4 to each side, therefore...
B= 28
A= 16.
Hope this helps!!
~Brainliest
Answer:
- <u>The sum of the series = 198.</u>
<u></u>
Step-by-step explanation:
In the given arithmetic series,
- The first term (a) = 6
- Common difference (d) = (aₙ – aₙ₋₁) = 10 - 6 = 4
- Last term (aₙ) = 38
To find the sum of the series, we need to find the number of terms (n) at first. So,
![a_{n} = a + (n - 1)d\\38 = 6 + (n - 1) 4\\38 - 6 = 4n - 4\\32 = 4n - 4\\32 + 4 = 4n\\36 = 4n\\36 \div 4 = n\\\boxed{9 = n}](https://tex.z-dn.net/?f=a_%7Bn%7D%20%3D%20a%20%2B%20%28n%20-%201%29d%5C%5C38%20%3D%206%20%2B%20%28n%20-%201%29%204%5C%5C38%20-%206%20%3D%204n%20-%204%5C%5C32%20%3D%204n%20-%204%5C%5C32%20%2B%204%20%3D%204n%5C%5C36%20%3D%204n%5C%5C36%20%5Cdiv%204%20%3D%20n%5C%5C%5Cboxed%7B9%20%3D%20n%7D)
Now, let's find the sum of the arithmetic series (Sₙ).
![S_{n} = \frac{n}{2} [2a + (n - 1)d]\\S_{n} = \frac{9}{2} [2*6 + (9 -1)4]\\S_{n} = \frac{9}{2} [12+ (8*4)]\\S_{n} = \frac{9}{2} [12+ 32]\\S_{n} = \frac{9}{2} (44)\\S_{n} = 9*22\\\boxed{S_{n} = 198}](https://tex.z-dn.net/?f=S_%7Bn%7D%20%3D%20%5Cfrac%7Bn%7D%7B2%7D%20%5B2a%20%2B%20%28n%20-%201%29d%5D%5C%5CS_%7Bn%7D%20%3D%20%5Cfrac%7B9%7D%7B2%7D%20%5B2%2A6%20%2B%20%289%20-1%294%5D%5C%5CS_%7Bn%7D%20%3D%20%5Cfrac%7B9%7D%7B2%7D%20%5B12%2B%20%288%2A4%29%5D%5C%5CS_%7Bn%7D%20%3D%20%5Cfrac%7B9%7D%7B2%7D%20%5B12%2B%2032%5D%5C%5CS_%7Bn%7D%20%3D%20%5Cfrac%7B9%7D%7B2%7D%20%2844%29%5C%5CS_%7Bn%7D%20%3D%209%2A22%5C%5C%5Cboxed%7BS_%7Bn%7D%20%3D%20198%7D)
- The sum of the series = <u>198</u>.
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Hope it helps!
![\mathfrak{Lucazz}](https://tex.z-dn.net/?f=%5Cmathfrak%7BLucazz%7D)