Answer:
points are( 6,2),(7,5) and (3,8)
Step-by-step explanation:
X,y changes into y,-X
Add (7x^2-2x) + (4x-3)
1 answer:
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First find out the function for the given table.
Find the slope.
m= (y2-y1)/(x2-x1)
m=(-1-(-3)) / (-1-(-2))
m=(-1+3) / (-1+2)
m=2/1
m=2
Now equation of line:
y-y1=m(x-x1)
y-(-3)=2(x-(-2))
y+3=2(x+2)
y+3=2x+4
y=2x+1
Now plot these points on the graph, we get the attached graph.
Find the slope.
m= (y2-y1)/(x2-x1)
m=(-1-(-3)) / (-1-(-2))
m=(-1+3) / (-1+2)
m=2/1
m=2
Now equation of line:
y-y1=m(x-x1)
y-(-3)=2(x-(-2))
y+3=2(x+2)
y+3=2x+4
y=2x+1
Now plot these points on the graph, we get the attached graph.
The correct answer is: Answer choice: [A]:
__________________________________________________________
→ "
" ; " { u 
± 3 } " ;
→ or, write as: " u / (u − 3) " ; {" u ≠ 3 "} AND: {" u ≠ -3 "} ;
__________________________________________________________
Explanation:
__________________________________________________________
We are asked to simplify:
;
Note that the "numerator" —which is: "(u² + 3u)" — can be factored into:
→ " u(u + 3) " ;
And that the "denominator" —which is: "(u² − 9)" — can be factored into:
→ "(u − 3) (u + 3)" ;
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Let us rewrite as:
___________________________________________________________
→
;
___________________________________________________________
→ We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ; since:
"
" ;
→ And we have:
_________________________________________________________
→ " " ; that is: " u / (u − 3) " ; { u
} .
and: { u
} .
→ which is: "Answer choice: [A] " .
_________________________________________________________
NOTE: The "denominator" cannot equal "0" ; since one cannot "divide by "0" ;
and if the denominator is "(u − 3)" ; the denominator equals "0" when "u = -3" ; as such:
"u3" ;
→ Note: To solve: "u + 3 = 0" ;
Subtract "3" from each side of the equation;
→ " u + 3 − 3 = 0 − 3 " ;
→ u = -3 (when the "denominator" equals "0") ;
→ As such: " u -3 " ;
Furthermore, consider the initial (unsimplified) given expression:
→ ;
Note: The denominator is: "(u² − 9)" .
The "denominator" cannot be "0" ; because one cannot "divide" by "0" ;
As such, solve for values of "u" when the "denominator" equals "0" ; that is:
_______________________________________________________
→ " u² − 9 = 0 " ;
→ Add "9" to each side of the equation ;
→ u² − 9 + 9 = 0 + 9 ;
→ u² = 9 ;
Take the square root of each side of the equation;
to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ;
→ √(u²) = √9 ;
→ | u | = 3 ;
→ " u = 3" ; AND; "u = -3 " ;
We already have: "u = -3" (a value at which the "denominator equals "0") ;
We now have "u = 3" ; as a value at which the "denominator equals "0");
→ As such: " u" ; "u -3 " ;
or, write as: " { u ± 3 } " .
_________________________________________________________
__________________________________________________________
→ "
→ or, write as: " u / (u − 3) " ; {" u ≠ 3 "} AND: {" u ≠ -3 "} ;
__________________________________________________________
Explanation:
__________________________________________________________
We are asked to simplify:
Note that the "numerator" —which is: "(u² + 3u)" — can be factored into:
→ " u(u + 3) " ;
And that the "denominator" —which is: "(u² − 9)" — can be factored into:
→ "(u − 3) (u + 3)" ;
___________________________________________________________
Let us rewrite as:
___________________________________________________________
→
___________________________________________________________
→ We can simplify by "canceling out" BOTH the "(u + 3)" values; in BOTH the "numerator" AND the "denominator" ; since:
"
→ And we have:
_________________________________________________________
→ "
and: { u
→ which is: "Answer choice: [A] " .
_________________________________________________________
NOTE: The "denominator" cannot equal "0" ; since one cannot "divide by "0" ;
and if the denominator is "(u − 3)" ; the denominator equals "0" when "u = -3" ; as such:
"u
→ Note: To solve: "u + 3 = 0" ;
Subtract "3" from each side of the equation;
→ " u + 3 − 3 = 0 − 3 " ;
→ u = -3 (when the "denominator" equals "0") ;
→ As such: " u
Furthermore, consider the initial (unsimplified) given expression:
→
Note: The denominator is: "(u² − 9)" .
The "denominator" cannot be "0" ; because one cannot "divide" by "0" ;
As such, solve for values of "u" when the "denominator" equals "0" ; that is:
_______________________________________________________
→ " u² − 9 = 0 " ;
→ Add "9" to each side of the equation ;
→ u² − 9 + 9 = 0 + 9 ;
→ u² = 9 ;
Take the square root of each side of the equation;
to isolate "u" on one side of the equation; & to solve for ALL VALUES of "u" ;
→ √(u²) = √9 ;
→ | u | = 3 ;
→ " u = 3" ; AND; "u = -3 " ;
We already have: "u = -3" (a value at which the "denominator equals "0") ;
We now have "u = 3" ; as a value at which the "denominator equals "0");
→ As such: " u
or, write as: " { u
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