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)" ;
___________________________________________________________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:
"u

3" ;
→ 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 } " .
_________________________________________________________
Hello,
Very nice as problem.
2 solutions:
1 quater,8 dimes, 2 pennies
and
3 quaters,3 dimes, 2 pennies
since
107=( 0, 0, 107) but : 100= 0*25+ 0*10+ 100
107=( 0, 1, 97) but : 100= 0*25+ 1*10+ 90
107=( 0, 2, 87) but : 100= 0*25+ 2*10+ 80
107=( 0, 3, 77) but : 100= 0*25+ 3*10+ 70
107=( 0, 4, 67) but : 100= 0*25+ 4*10+ 60
107=( 0, 5, 57) but : 100= 0*25+ 5*10+ 50
107=( 0, 6, 47) but : 100= 0*25+ 6*10+ 40
107=( 0, 7, 37) but : 100= 0*25+ 7*10+ 30
107=( 0, 8, 27) but : 100= 0*25+ 8*10+ 20
107=( 0, 9, 17) but : 100= 0*25+ 9*10+ 10
107=( 0, 10, 7) but : 100= 0*25+ 10*10+ 0
107=( 1, 0, 82) but : 100= 1*25+ 0*10+ 75
107=( 1, 1, 72) but : 100= 1*25+ 1*10+ 65
107=( 1, 2, 62) but : 100= 1*25+ 2*10+ 55
107=( 1, 3, 52) but : 100= 1*25+ 3*10+ 45
107=( 1, 4, 42) but : 100= 1*25+ 4*10+ 35
107=( 1, 5, 32) but : 100= 1*25+ 5*10+ 25
107=( 1, 6, 22) but : 100= 1*25+ 6*10+ 15
107=( 1, 7, 12) but : 100= 1*25+ 7*10+ 5
107=( 1, 8, 2) is good
107=( 2, 0, 57) but : 100= 2*25+ 0*10+ 50
107=( 2, 1, 47) but : 100= 2*25+ 1*10+ 40
107=( 2, 2, 37) but : 100= 2*25+ 2*10+ 30
107=( 2, 3, 27) but : 100= 2*25+ 3*10+ 20
107=( 2, 4, 17) but : 100= 2*25+ 4*10+ 10
107=( 2, 5, 7) but : 100= 2*25+ 5*10+ 0
107=( 3, 0, 32) but : 100= 3*25+ 0*10+ 25
107=( 3, 1, 22) but : 100= 3*25+ 1*10+ 15
107=( 3, 2, 12) but : 100= 3*25+ 2*10+ 5
107=( 3, 3, 2) is good
107=( 4, 0, 7) but : 100= 4*25+ 0*10+ 0
Hello, Oliveria. I am saddened to hear that you have started a whole rewrite of your plan for mathematics, but fear not; there are a couple of resources and websites that are always ready for you. For starters, visit websites that have at least 7th or 6th grade and up in benchmark mathematics content; this would be IXL, KhanAcademy, or even take some of the lesson tutorials from places like Symbolab and mathpoppa (<span>please, don't use the algebra calculator a lot. It will add more to your demise). Any how, you can always ask your guidance counselor at your school for help, or even your algebra teacher for some tutorials on certain topics. Always at least try to ask for help from your teacher or friends; even if you think they can't help much, it is always best to try
It seems that you have enough self direction and initiative to go on this route, now you just need the additional boosts :)</span><span />
Answer:
significance level
Step-by-step explanation:
In other to determine whether her sample proportion is "different enough" than the claimed population proportion, We will set significance level because p-value should be compared to significance level that will allow Joanna to make a conclusion about whether her sample proportion is "different enough" than the claimed population proportion.