H = 3b+2
A = (h*b)/2 28 = (3b+2)b/2 56 = 3b²+2b 0 = 3b² + 2b - 56
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![\left \{ {{y=2} \atop {x=2}} \right. \int\limits^a_b {x} \, dx \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta \\ \\ \\ x^{2} \sqrt{x} \sqrt[n]{x} \frac{x}{y} x_{123} x^{123} \leq \geq \pi \alpha \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] x_{123} \int\limits^a_b {x} \, dx \left \{ {{y=2} \atop {x=2}}](https://tex.z-dn.net/?f=%20%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D%20%5Cright.%20%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%20%5Cbeta%20%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%20x%5E%7B2%7D%20%20%5Csqrt%7Bx%7D%20%20%5Csqrt%5Bn%5D%7Bx%7D%20%20%5Cfrac%7Bx%7D%7By%7D%20%20x_%7B123%7D%20%20x%5E%7B123%7D%20%20%5Cleq%20%20%5Cgeq%20%20%5Cpi%20%20%5Calpha%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%20x_%7B123%7D%20%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20%20%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D)
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I think 12 might be the second one, I'm not sure though!
Answer:
82
Step-by-step explanation:
Well since everything is already in order from least to greatest:
48, 56, 68, 72, 72, 78, 78, 80, 82, 84, 88, 88, 88, 90, 94, 98, 100
The median is the number in the middle
So 82 is the median
Hope this helps!
Feel free to comment any questions you have!
The initial statement is: QS = SU (1)
QR = TU (2)
We have to probe that: RS = ST
Take the expression (1): QS = SU
We multiply both sides by R (QS)R = (SU)R
But (QS)R = S(QR) Then: S(QR) = (SU)R (3)
From the expression (2): QR = TU. Then, substituting it in to expression (3):
S(TU) = (SU)R (4)
But S(TU) = (ST)U and (SU)R = (RS)U
Then, the expression (4) can be re-written as:
(ST)U = (RS)U
Eliminating U from both sides you have: (ST) = (RS) The proof is done.
