The distance between the two points pictured is d = Vn
1 answer:
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begin by pulling 2 out of the numerator using the distributive property.
numerator: 2(16x^4 - 25) and now factor
numerator: 2(4x^2 - 5)(4x^2 + 5)
Now go to the denominator. It looks messy but it will break down.
Pull out 4x^2 for the first two terms and 5 for the last 2 terms. Use the distributive property.
denominator: 4x^2(x - 3) - 5(x - 3) now x - 3 is the common term.
denominator: (x - 3)(4x^2 - 5)
Put the two results together.
After canceling out 4x^2 - 5 on both the numerator and the denominator, you are left with.
Answer:
The fraction of the applicants who have passed the two phases of the selection process is .
Step-by-step explanation:
The question is:
In the first phase of the selection process for the body of teachers they eliminate four-ninth of the applicants while in the second phase they eliminate 5/7 of those who passed the first phase. If there were 315 applicants what is the fraction of the applicants who have passed the two phases of the selection process?
The total number of applicants is
<em>N</em> = 315.
It is provided that 4/9th of the applicants were eliminated in the first phase.
Then the fraction of candidates who passed the first phase is:
Fraction who passed phase I =
Now, it is also provided that 5/7 of those who passed the first phase were eliminated in the second phase.
Then the fraction of candidates who passed the second phase is:
Fraction who passed phase II =
That is, 2/7 of the remaining applicants passed the second phase.
The fraction of the applicants who have passed the two phases is:
Thus, the fraction of the applicants who have passed the two phases of the selection process is .