The missing provided information is that Mr. Nicholson accepts a job that pays an annual salary of
$60,000. And he is given the option of choosing between two annual raises:
a) an annual raise of $3,500 or b) an annual raise of 5% of his current salary.
Then, with that information you have to answer the given questions.Which I am going to do step by step.
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1) identify each of Mr. Nicholson’s earning opportunities as arithmetic or
geometric. For each opportunity, include the common difference or
ratio. In your final answer, use complete sentences to explain how you
identified each opportunity as arithmetic or geometric.
- An annual raise of fix $3500 means that every year the salary increase in a constant amount driving to this sequence:
60,000 + 3500 = 63,500;
63,500 + 3,500 = 67,000
67,000 + 3,500 = 70,500
70,500 + 3,500 = 74,000
74,000 + 3,500 = 77,500
...
Then you have a constant difference between two adjacent terms which means that this is an arithmethic progression.
- An annual raise of 5% of the current salary, means that the salary will increase by a constant factor of 1.05, driving to this sequence:
60,000 * 1.05 = 63,000
63,000 * 1.05 = 66,150
66,150 * 1.05 = 69,457.50
69,457.50 * 1.05 = 72,930.375
72,930.375 * 1.05 = 76,576.89
...
In this case, the increase is geometrical because you have that two adjacent terms differentiate by a constant factor, e.g.: 69,457.50 / 66,150 = 1.50.
2)Model each of Mr. Nicholson’s salary options with a recursive sequence
that includes his potential earnings for the first three years of
employment.
According to the first three terms of each sequence, can you conclude
that there is a significant difference in Mr. Nicholson’s potential
earnings with each increase option? Use complete sentences to explain
your conclusion.
Models
- Atrihmetic progression option
Annual salary the year n= Sn
Initial Salary = S1 = 60,000
difference, d = 3500
number of year: n
Model: S = S1 + (n-1)*d
S = 60,000 + (n-1)*3500
Potential earnings for first three years:
You can use the fomula for the sum of n terms in an arithmetic progression: [S1 + S3]*(n) / (2)
Sum = [60,000 + 67000] * 3 / 2 = 190,500
This is the same that [60,000 + 63,500 + 67,000] = 190,500.
- Geometric progression:
S1 = 60,000
r = 1.05
Sn = S1 * r^(n-1) = 60,000 - (1.05)^(n-1)
Potential earnings the first three years:
60,000 + 63,000 + 66,150 = 189,150
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Now you got that there is a substantial difference in potential earnings with each option: the constant increase of $3500 (arithmetic progression) during three years results in a bigger earning for that time, because 5% of difference the second year is only 3000, and the third year 3150; both below $3500. This results in that the arithmetic progression is better for Mr. Nicholson during the first three years.
Answer: y = 2x^4
The graph represents a parabolic-looking shape so it's either a quadratic or some even degree polynomial. This rules out y = -2x^3 and y = 2x^3 as they are cubics with odd degrees
The degree of a polynomial is the largest exponent. It determines the end behavior. In this case, both ends are rising upward together. Because they are rising up in the positive y direction, this means that we cannot have y = -2x^4 as the answer. For example, if x = 2 then y = -2*x^4 = -2*2^4 = -32, but the point (x,y) = (2,-32) isn't on the blue curve. So we can rule y = -2x^4 out.
The fact that y = 2x^4 has a positive leading coefficient tells us that the endpoints point upward.
Answer:
The ratio of perimeter of ABCD to perimeter of WXYZ = 
Step-by-step explanation:
First, we have to determine the multiplicative factor of the dimensions for both figures.
Considering sides AB and WX,
multiplicative factor = 
= 1.5
So that:
XY = 6 x 1.5 = 9
YZ = 7 x 1.5 = 10.5
ZW = 7 x 1.5 = 10.5
Perimeter of ABCD = 6 + 7 + 7 + 8
= 28
Perimeter of WXYZ = 9 + 10.5 + 10.5 + 12
= 42
The ratio of the perimeters of the two quadrilaterals can be determined as;
ratio = 
= 
=
The ratio of the perimeter of ABCD to perimeter of WXYZ is .
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
