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Answer:
9 pitchers
Step-by-step explanation:
Given
Cylinder 1:
Cylinder 2:
Required
How many pitchers' cylinder 2 can fill
First, we calculate the volume of both cylinders
Volume is calculated as:
For A:
For B:
In (a):
So, we have ve:
<em>If the first cylinder can fill 1, then the second can fill 9 pitchers</em>
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:
"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
_________________________________________________________
Average speed =
Ishan's Speed uphill = 10mph
Time taken = 18 minutes = 18/60 = 3/10 hours
Now distance = rate × time
Distance traveled by Ishaan up hill = 10 × 3/10 = 3 miles
Now returning back home
Distance = 3 miles
Rate = 30 mph
So time taken = distance/ rate = 3/30= 1/10 hours = 1/10 ×60 = 6 minutes.
Now total distance traveled = 3 + 3 = 6 miles
Total time taken = 18 + 6 = 24 minutes = 24/60 = 2/5 hours
Average speed
=
= 15 mph
Ishaan's average speed for the entire trip from home to the gift store and back is 15mph