Answer:
so idc![\sqrt[n]{x} \sqrt{x} \alpha \pi x^{2} \\ \left \{ {{y=2} \atop {x=2}} \right. x_{123} \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]](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%7D%20%5Csqrt%7Bx%7D%20%5Calpha%20%5Cpi%20x%5E%7B2%7D%20%5C%5C%20%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D%20%5Cright.%20x_%7B123%7D%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D)
443
Step-by-step explanation: its 2 6\7
Answer:
im not 100% sure but i pretty sure its 2.
Step-by-step explanation:
To be honest, I'm not sure which four steps your teacher is referring to. However, I'll show you one way to graph this.
A graph is simply a collection of points. Often those points are connected in some way (though they don't necessarily have to be) to form a curve.
Each point is of the form (x,y). To get each point, we pick random x values and determine their paired y value counterpart.
For example, if we pick x = -3, then,
y= -x^2 -4x -3
y= -(-3)^2 -4(-3) -3
y = -9 - 4(-3) - 3
y = -9 + 12 - 3
y = 0
This indicates that (-3, 0) is one point on the curve.
Let's repeat for x = -2
y= -x^2 -4x -3
y= -(-2)^2 -4(-2) -3
y = -4 - 4(-2) - 3
y = -4 + 8 - 3
y = 1
So (-2, 1) is another point on the curve.
Repeat this process as many times as you want. You should do at least 3 or 4 points in my opinion. The more points you generate, the more accurate the curve. After generating the points, you'll plot them all on the same xy grid. Then finally draw a curve through all of the points as shown below.
I used GeoGebra to make the graph.