45 x 5 = 225 so 225 students altogether on 5 buses
Answer: x = 1
Step-by-step explanation:
substitute 1 for x
1/2(10(1)+6) = -2(1)+10
1/2(10+6) = -2+10
5+3 = -2+10
8=8
Answer:
-2
Step-by-step explanation:
Start with the vertex form of the equation of a parabola: y - k = a(x - h)^2
Here h = -2, k = -3, x = -1, y = -5. Find a:
-5 - [-3] = a(-1 - [-2])^2, or
-5 + 3 = a(1)^2, or
-2 = a
The unknown coefficient is -2.
Answer:
he gets 104$
Step-by-step explanation:
104 dollars
26x4=104
Answer:


Step-by-step explanation:
<u>Taylor series</u> expansions of f(x) at the point x = a

This expansion is valid only if
exists and is finite for all
, and for values of x for which the infinite series converges.






Substituting the values in the series expansion gives:

Factoring out e⁴:
![e^{4x}=e^4\left[1+4(x-1)+8}(x-1)^2+...\right]](https://tex.z-dn.net/?f=e%5E%7B4x%7D%3De%5E4%5Cleft%5B1%2B4%28x-1%29%2B8%7D%28x-1%29%5E2%2B...%5Cright%5D)
<u>Taylor Series summation notation</u>:

Therefore:
