Let x and y be the 2 parts of 15 ==> x + y=15 (given)
Reciprocal of x and y ==> 1/x +1/y ==> 1/x + 1/y = 3/10 (given)
Let's solve 1/x + 1/y = 3/10 . Common denominator = 10.x.y (reduce to same denominator)
==> (10y+10x)/10xy = 3xy/10xy ==> 10x+10y =3xy
But x+y = 15 , then 10x+10y =150 ==> 150=3xy and xy = 50
Now we have the sum S of the 2 parts that is S = 15 and
their Product = xy =50
Let's use the quadratic equation for S and P==> X² -SX +P =0
Or X² - 15X + 50=0, Solve for X & you will find:
The 1st part of 15 is 10 & the 2nd part is 5
Answer:
Huh
Step-by-step explanation:
Answer:
no
You said skip the explanation, so ok
As written, the denominator in both fractions is x, so the only restriction on the domain is ... x ≠ 0.
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We suspect you intend ...
... f(x) = 2/(x-4) +1/(x+2)
which is undefined when x = 4 or x = -2.
The domain is all real numbers except -2 and 4.
Well, she could do 30/10 which equals 3. So she could do 10 groups with 3 students in each group.
