Answer:
plz if you knew if you new it plz tell me
Answer:
I think that the required options are wrong.
Answer is given below with explanations.
mZCBD = 41 degrees
Step-by-step explanation:
![from \: the \: \: above \: mentioned \: diagram \\ we \: can \: say \: that \: \\( angle \: ABD) + (angle \: DBC) = 68 \\ given \: that \: \: angle \: ABD = 2x + 5 \\ and \: \: angle \: CBD = 3x + 8 \\ on \: substituting \: the \: values \\ (2x + 5) + (3x + 8) = 68 \\ 5x + 13 = 68 \\ 5x = 68 - 13 \\ 5x = 55 \\ x = \frac{55}{5} \\ x = 11 \\ angle \: CBD = 3x + 8 \\ \: \: = 3(11) + 8 \\ \: \: = 33 + 8 \\ \: \: angle \: CBD= 41](https://tex.z-dn.net/?f=from%20%5C%3A%20the%20%5C%3A%20%20%5C%3A%20above%20%5C%3A%20mentioned%20%5C%3A%20diagram%20%5C%5C%20we%20%5C%3A%20can%20%5C%3A%20say%20%5C%3A%20that%20%5C%3A%20%20%5C%5C%28%20angle%20%5C%3A%20ABD%29%20%2B%20%28angle%20%5C%3A%20DBC%29%20%3D%2068%20%5C%5C%20given%20%5C%3A%20that%20%5C%3A%20%20%5C%3A%20angle%20%5C%3A%20ABD%20%3D%202x%20%2B%205%20%5C%5C%20and%20%5C%3A%20%5C%3A%20%20angle%20%5C%3A%20CBD%20%3D%203x%20%2B%208%20%5C%5C%20on%20%5C%3A%20substituting%20%5C%3A%20the%20%5C%3A%20values%20%5C%5C%20%282x%20%2B%205%29%20%2B%20%283x%20%2B%208%29%20%3D%2068%20%5C%5C%205x%20%2B%2013%20%3D%2068%20%5C%5C%205x%20%3D%2068%20-%2013%20%5C%5C%205x%20%3D%2055%20%5C%5C%20x%20%3D%20%20%5Cfrac%7B55%7D%7B5%7D%20%20%5C%5C%20x%20%3D%2011%20%5C%5C%20angle%20%5C%3A%20CBD%20%3D%203x%20%2B%208%20%5C%5C%20%20%5C%3A%20%20%5C%3A%20%20%3D%203%2811%29%20%2B%208%20%5C%5C%20%20%5C%3A%20%20%5C%3A%20%20%3D%2033%20%2B%208%20%5C%5C%20%20%5C%3A%20%20%5C%3A%20%20angle%20%5C%3A%20CBD%3D%2041)
<em>HAVE A NICE DAY</em><em>!</em>
<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>
Answer:
It's the last answer choice.
Step-by-step explanation:
We need to solve for b, so divide each side by H, and then multiply each side by 2 to get rid of the 1/2
I had a whole answer and my tab crashed. Hate that.
We'll use the known sum of the first n natural numbers,
![\displaystyle \sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csum_%7Bk%3D1%7D%5E%7Bn%7D%20k%20%3D%20%5Cdfrac%7Bn%28n%2B1%29%7D%7B2%7D)
Let a be the first number, the one we seek.
![\displaystyle 3^9= \sum_{k=0}^{26} (a+2k) = 27a+ 2 \sum_{k=1}^{26}k](https://tex.z-dn.net/?f=%5Cdisplaystyle%203%5E9%3D%20%5Csum_%7Bk%3D0%7D%5E%7B26%7D%20%28a%2B2k%29%20%3D%2027a%2B%202%20%5Csum_%7Bk%3D1%7D%5E%7B26%7Dk)
![\displaystyle 3^9=27a+2(26)(26+1)/2](https://tex.z-dn.net/?f=%5Cdisplaystyle%203%5E9%3D27a%2B2%2826%29%2826%2B1%29%2F2%20)
![3^9=27(a+26)](https://tex.z-dn.net/?f=3%5E9%3D27%28a%2B26%29%20)
![3^6= a+ 26](https://tex.z-dn.net/?f=3%5E6%3D%20a%2B%2026)
![a = 3^6 - 26](https://tex.z-dn.net/?f=a%20%3D%203%5E6%20-%2026)
![a = 703 \quad\checkmark](https://tex.z-dn.net/?f=a%20%3D%20703%20%5Cquad%5Ccheckmark)
Answer:
197.50
Step-by-step explanation: