Answer:
nickels 15 dimes 45 pennies and quarters 10
F(x)=(-2/((x+y-2)^(1/2))-(x+y+2)^(1/2)
the only irrational part of this expression is the (x+y-2)^(1/2) in the denominator, so, to rationalize this, you multiply the numerator and denominator by the denominator, as well as the other parts of the expression
also, you must multiply the -sqrt(x+y+2) by sqrt(x+y-2)/sqrt(x+y-2) to form a common denominator
(-2)/(x+y-2)^(1/2)-(x+y+2)^(1/2)(x+y-2)^(1/2)/(x+y-2)^(1/2)
(common denominator)
(-2-(x^2+xy+2x+xy+y^2+2y-2x-2y-4))/(x+y-2)^(1/2)
(FOIL)
(-2-x^2-y^2-2xy+4)/(x+y-2)^(1/2)
(Distribute negative)
(-x^2-y^2-2xy+2)/(x+y-2)^(1/2)
(Simplify numerator)
(-x^2-y^2-2xy+2)(x+y-2)^(1/2)/(x+y-2)^(1/2)(x+y-2)^(1/2)
(Rationalize denominator by multiplying both top and bottom by sqrt)
(-x^2-y^2-2xy+2)((x+y-2)^(1/2))/(x+y-2)
(The function is now rational)
=(-x^2-y^2-2xy+2)(sqrt(x+y-2))/(x+y-2)
Answer:
first we find the common difference.....do this by subtracting the first term from the second term. (9 - 3 = 6)...so basically, ur adding 6 to every number to find the next number.
we will be using 2 formulas....first, we need to find the 34th term (because we need this term for the sum formula)
an = a1 + (n-1) * d
n = the term we want to find = 34
a1 = first term = 3
d = common difference = 6
now we sub
a34 = 3 + (34-1) * 6
a34 = 3 + (33 * 6)
a34 = 3 + 198
a34 = 201
now we use the sum formula
Sn = (n (a1 + an)) / 2
S34 = (34(3 + 201))/2
s34 = (34(204)) / 2
s34 = 6936/2
s34 = 3468 <=== the sum of the first 34 terms:
There are 12 student(s) in the group, because there is 12 one-third part(s) in 4 while jars
Hi there.
The answer is definitely "D" because on the graph it shows there are only 29 students that did the test.
