How much larger is the sum of 3x²-2x-8 and 4x²+2x-1 than the sum of 5x²-4x and 2x²-13x+4
1 answer:
You might be interested in
Answer:
Problem 2):
which agrees with answer C listed.
Problem 3) : D = (-3, 6] and R = [-5, 7]
which agrees with answer D listed
Step-by-step explanation:
Problem 2)
The Domain is the set of real numbers in which the function (given by a graph in this case) is defined. We see from the graph that the line is defined for all x values between 0 and 240. Such set, expressed in "set builder notation" is:
Problem 3)
notice that the function contains information on the end points to specify which end-point should be included and which one should not. The one on the left (for x = -3 is an open dot, indicating that it should not be included in the function's definition, therefor the Domain starts at values of x strictly larger than -3. So we use the "parenthesis" delimiter in the interval notation for this end-point. On the other hand, the end point on the right is a solid dot, indicating that it should be included in the function's definition, then we use the "square bracket notation for that end-point when writing the Domain set in interval notation:
Domain = (-3, 6]
For the Range (the set of all those y-values connected to points in the Domain) we use the interval notation form:
Range = [-5, 7]
since there minimum y-value observed for the function is at -5 , and the maximum is at 7, with a continuum in between.
Answer:
1i) nth term = 12(-b/4)^(n-1)
ii) nth term = 3(-1/9)^(n-1)
2i) Sixth term= (-3b^5)/256
ii) Eight term = -1/1594323
Question:
The complete question as found on brainly (question-10746111):
Write down the nth term of each of the following G.Ps whose first two terms are given as follows. Also find the term stated besides each G.P.
i) 12,-3b,......sixth term
ii) 3,-1/3,..........;8th term?
Step-by-step explanation:
1) To solve terms involving Geometric progressions (GPs), we would first state the variables that are to be incorporated in the formula.
i) 12,-3b,......;sixth term
1st term = a= 12
r = common ratio = 2nd term /1st term
r = -3b/12 = -b/4
Then we would find the nth term
See attachment for details
ii) 3,-1/3,..........;8th term
1st term = a= 3
r = common ratio = 2nd term /1st term
r = (-⅓)/3 = (-⅓)(⅓) = -1/9
Then we would find the nth term
See attachment for details
2) we are to determine the sixth term in (i) and eight term in (ii)
See attachment for details
Sixth term= (-3b^5)/256
Eight term = -1/1594323
