Answer:
Check the explanation
Step-by-step explanation:
in summarizing the findings at the end. with (a) In=1(-1)"sin () (-1)" (n-2) (C) 2n=1 ºnin (c) En=1 (e) En=4 (-1)", (b) 20 (-1)" (n2+n+ 2) (0) 2n=1 n42) (d) =2 (–1)M (1+2)" (1) =1 (–1)n+1 2012 4+ 2n+ 3 4) Consider the series En
As well as to calculate the smallest integer, The negative integers will go on with no bound and getting lesser and lesser for ever.
Kindly check the attached images for the full step by step explanation to the question.
Both <em>Px</em> + <em>Qy</em> = <em>R</em> and <em>Tx</em> + <em>Uy</em> = <em>V</em> have the same solution (2, 9), meaning the point (2, 9) lies on both lines such that <em>x</em> = 2 and <em>y</em> = 9 makes both equations true.
When you add these equations together, you get
(<em>Px</em> + <em>Qy</em>) + (<em>Tx</em> + <em>Uy</em>) = <em>R</em> + <em>V</em>
→ (<em>P</em> + <em>T</em> ) <em>x</em> + (<em>Q</em> + <em>U</em> ) <em>y</em> = <em>R</em> + <em>V</em>
so the first option is equivalent.
When you subtract the first equation from the second, you get
(<em>Tx</em> + <em>Uy</em>) - (<em>Px</em> + <em>Qy</em>) = <em>V</em> - <em>R</em>
→ (<em>T</em> - <em>P</em>) <em>x</em> + (<em>U</em> - <em>Q</em>) <em>y</em> = <em>V</em> - <em>R</em>
so the third option is also equivalent.
Answer:
b) 1
Step-by-step explanation:
First, bring the -3 to the right side.
x²+4x + 4
now use b²-4ac.
4²-4(1)(4)
16-16
0
If discriminant < 0 there are 2 imaginary solutions
if discriminant > 0 there are 2 real solutions
if discriminant = 0 there is 1 real solution
The discriminant here is 0 so it has one real solution
1 6 tenths 3 hundredths 9 thousandths
G
2^2-7
4-7
-3
f
2(-3)+1
-5+1
-4 is your answer