By definition of supplementary angles,
∠ABD + ∠BDE = 180
5x + 17x - 18 = 180
22x = 198
x = 9
∠ABD = 5(9) = 45°
∠BDE = 180 - 45 = 135°
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❖ The median of the data is 23.
You keep crossing out numbers until you get to the middle point. First, you cross of the beginning and the end which is 21 and 25. You then cross out 22 and 24. The remaining number is 23. When you are working with an even amount of numbers, you are left with 2 numbers so you add those 2 numbers and divide by 2.
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~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ
the way I get the subsequent term, nevermind the exponents, the exponents part is easy, since one is decreasing and another is increasing, but the coefficient, to get it, what I usually do is.
multiply the current coefficient by the exponent of the first-term, and divide that by the exponent of the second-term + 1.
so if my current expanded term is say 7a³b⁴, to get the next coefficient, what I do is (7*3)/5 <----- notice, current coefficient times 3 divided by 4+1.
anyhow, with that out of the way, lemme proceed in this one.

so, following that to get the next coefficient, we get those equivalents as you see there for the 2nd and 3rd terms.
so then, we know that the expanded 2nd term is 24x therefore

we also know that the expanded 3rd term is 240x², therefore we can say that

but but but, we know what "n" equals to, recall above, so let's do some quick substitution
![\bf a^2n^2-a^2n=480\qquad \boxed{n=\cfrac{24}{a}}\qquad a^2\left( \cfrac{24}{a} \right)^2-a^2\left( \cfrac{24}{a} \right)=480 \\\\\\ a^2\cdot \cfrac{24^2}{a^2}-24a=480\implies 24^2-24a=480\implies 576-24a=480 \\\\\\ -24a=-96\implies a=\cfrac{-96}{-24}\implies \blacktriangleright a = 4\blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ n=\cfrac{24}{a}\implies n=\cfrac{24}{4}\implies \blacktriangleright n=6 \blacktriangleleft](https://tex.z-dn.net/?f=%5Cbf%20a%5E2n%5E2-a%5E2n%3D480%5Cqquad%20%5Cboxed%7Bn%3D%5Ccfrac%7B24%7D%7Ba%7D%7D%5Cqquad%20a%5E2%5Cleft%28%20%5Ccfrac%7B24%7D%7Ba%7D%20%5Cright%29%5E2-a%5E2%5Cleft%28%20%5Ccfrac%7B24%7D%7Ba%7D%20%5Cright%29%3D480%20%5C%5C%5C%5C%5C%5C%20a%5E2%5Ccdot%20%5Ccfrac%7B24%5E2%7D%7Ba%5E2%7D-24a%3D480%5Cimplies%2024%5E2-24a%3D480%5Cimplies%20576-24a%3D480%20%5C%5C%5C%5C%5C%5C%20-24a%3D-96%5Cimplies%20a%3D%5Ccfrac%7B-96%7D%7B-24%7D%5Cimplies%20%5Cblacktriangleright%20a%20%3D%204%5Cblacktriangleleft%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20n%3D%5Ccfrac%7B24%7D%7Ba%7D%5Cimplies%20n%3D%5Ccfrac%7B24%7D%7B4%7D%5Cimplies%20%5Cblacktriangleright%20n%3D6%20%5Cblacktriangleleft)
Answer:
+5
Step-by-step explanation:
the average rate of change is adding 5