The answer is B
Because if you use your calculation right you would end up with 16,463
the construction of fields of formal infinite series in several variables, generalizing the classical notion of formal Laurent series in one variable. Our discussion addresses the field operations for these series (addition, multiplication, and division), the composition, and includes an implicit function theorem.
(PDF) Formal Laurent series in several variables. Available from: https://www.researchgate.net/publication/259130653_Formal_Laurent_series_in_several_variables [accessed Oct 08 2018].
Hello,
Use the factoration
a^2 - b^2 = (a - b)(a + b)
Then,
x^2 - 81 = x^2 - 9^2
x^2 - 9^2 = ( x - 9).(x + 9)
Then,
Lim (x^2- 81) /(x+9)
= Lim (x -9)(x+9)/(x+9)
Simplity x + 9
Lim (x -9)
Now replace x = -9
Lim ( -9 -9)
Lim -18 = -18
_______________
The second method without using factorization would be to calculate the limit by the hospital rule.
Lim f(x)/g(x) = lim f(x)'/g(x)'
Where,
f(x)' and g(x)' are the derivates.
Let f(x) = x^2 -81
f(x)' = 2x + 0
f(x)' = 2x
Let g(x) = x +9
g(x)' = 1 + 0
g(x)' = 1
Then the Lim stay:
Lim (x^2 -81)/(x+9) = Lim 2x /1
Now replace x = -9
Lim 2×-9 = Lim -18
= -18
Answer:
<h2>
Check A, B and D </h2>
Step-by-step explanation:
A. m∠4 = 180° - m∠3
m∠1 = 180° - (m∠3 + m∠2)
m∠3 < m∠3 + m∠2 ⇒ m∠4 >m∠1
B. m∠4 = 180° - m∠3
m∠2 = 180° - (m∠3 + m∠1)
m∠3 < m∠3 + m∠1 ⇒ m∠4 >m∠2
C. Not necessarily {could be but not always}
D. m∠4 = 180° - m∠3
m∠1 + m∠2 + m∠3 = 180° ⇒ m∠1 + m∠2 = 180° - m∠3 = m∠4
E. m∠3 = m∠1 + m∠2 only if m∠3 = 90°
F. Not necessarily. {Could be but also could not be. In the picture is, but we don't know if the diagram is accurate or just approximate.}
Given three points
P1(-2,8)
P2(0,-4)
P3(4,68)
We need the quadratic equation that passes through all three points.
Solution:
We first assume the final equation to be
f(x)=ax^2+bx+c .............................(0)
Observations:
1. Points are not symmetric, so cannot find vertex visually.
2. Using the point (0,-4) we substitute x=0 into f(x) to get
f(0)=0+0+c=-4, hence c=-4.
3. We will use the two other points (P1 & P3) to set up a system of two equations to find a and b.
f(-2)=a(-2)^2+b(-2)-4=8 => 4a-2b-4=8.................(1)
f(4)=a(4^2)+b(4)-4=68 => 16a+4b-4=68.............(2)
4. Solve system
2(1)+(2) => 24a+0b-12=84 => 24a=96 => a=96/24 => a=4 ......(3)
substitute (3) in (2) => 16(4)+4b-4=68 => b=8/4 => b=2 ..........(4)
5. Put values c=-4, a=4, b=2 into equation (0) to get
f(x)=4x^2+2x-4
Check:
f(-2)=4((-2)^2)+2(-2)-4=16-4-4=8
f(0)=0+0-4 = -4
f(4)=4(4^2)+2(4)-4=64+8-4=68
So all consistent, => solution ok.
