H = 3b+2
A = (h*b)/2 28 = (3b+2)b/2 56 = 3b²+2b 0 = 3b² + 2b - 56
⊕![\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)
ω
l
∩
A = (h*b)/2 28 = (3b+2)b/2 56 = 3b²+2b 0 = 3b² + 2b - 56
⊕
ω
l
∩
Simplify. (-6u? - 9u-4)-(-5u²+2u+3)+(2u? + 6u+4)
1 answer:
You might be interested in
Answer:
see the explanation
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
m∠PQS+m∠SQR=m∠PQR ----> equation A (by Addition Angle Postulate)
we have that
m∠PQR=90° ----> equation B given problem (because is a right angle)
substitute equation B in equation A
m∠PQS+m∠SQR=90°
Remember that
Two angles re complementary is their sum is equal to 90 degrees (Definition of complementary angles)
therefore
m∠PQS and m∠SQR are complementary angles
