Answer: ![sds\\ \\ x^{2} \geq \int\limits^a_b {x} \, dx \lim_{n \to \infty} a_n \geq \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \pi \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \lim_{n \to \infty} a_n \int\limits^a_b {x} \, dx \left \{ {{y=2} \atop {x=2}} \right. x^{2} \lim_{n \to \infty} a_n \pi \neq \sqrt{x} \neq](https://tex.z-dn.net/?f=sds%5C%5C%20%5C%5C%20x%5E%7B2%7D%20%5Cgeq%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n%20%5Cgeq%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%5Cpi%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%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D%20%5Cright.%20x%5E%7B2%7D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n%20%5Cpi%20%5Cneq%20%5Csqrt%7Bx%7D%20%5Cneq)
Step-by-step explanation:i need the think points
The mess up occurs in one of the first things she says. She first claims that 7^2 +25^2 = 24^2
Wait a minute... slow your roll there.
25 is the hypotenuse, or the longest side of the triangle AND opposite the right angle (but we aren't supposed to know that yet).
The Pythagorean Theorem holds that a^2 + b^2 = c^2, where a & b are the legs of the triangle & c is the hypotenuse. So, she got the formatting correctly, but she switched 24 & 25.
Hope this helps!
Answer:
See explanation
Step-by-step explanation:
Jacob has 75 cookies.
He gave 2 cookies to each student.
Complete the table:

Let n be the number of students. After nth student arrived Jacob had left

cookies.
This is an arithmetic sequence with

Thus,

We know that the Circumcenter is the intersection of a triangle's right (perpendicular) bisectors. Once you have found two of these you can determine their point of intersection.
It is a proven mathematical concept that the third perpendicular bisector will also cross through this intersection point. While it is not necessary to construct all three it is often done so to verify that the intersection of the first two perpendicular bisectors was in fact correct.