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Anettt [7]
3 years ago
13

London was offered a job that paid a salary of \$93,000$93,000 in its first year. The salary was set to increase by 2% per year

every year. If London worked at the job for 22 years, what was the total amount of money earned over the 22 years, to the nearest whole number?
Mathematics
2 answers:
Mashcka [7]3 years ago
8 0

Answer: The total amount earned over the 22 year period would be $143,776

Step-by-step explanation: What the question simply requires is the value of the current amount (93000) at an annual increment of 2% (or 0.02) for a given number of years (22).

The formula for determining the amount is given as follows;

A = P(1 + r)∧t

Where A is the amount after the given period, P is the initial amount currently (93000), r is the rate of increase per annum (0.02) and t is the period in years (22). The formula now becomes;

A = 93000(1 + 0.02)²²

A = 93000(1.02)²²

A = 93000 x 1.54597967

A = 143776.10931

A ≈ 143776 (to the nearest whole number)

The total amount earned over the 22 year period would be $143,776

Tema [17]3 years ago
5 0

Answer: Amount = $143776

Step-by-step explanation:

Given that London was offered a job that paid a salary of $93,000 in its first year. The salary was set to increase by 2% per year every year.

Let P = 93000

Rate R = 2%

The amount of salary he will receive after 22 years can be calculated by using exponential equation

A = P(1 + 2%)^t

Where t = 22 years

A = 93000(1.02)^22

A = 93000 × 1.545979

A = 143776.11 dollars

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The fucntion f(x)=2x^2+3x+5 when evaluated, gives a value of 19. What is the functions input value?
BlackZzzverrR [31]

Answer:

x1= 14/4 and X2 = -2

Step-by-step explanation:

To find the values of "X" for which when the function is evaluated gives 19 we need to equal the function to 19, as follows:

2x^2+3x+5 = 19

2x^2+3x+5-19=0

2x^2+3x-14=0

Then, solving for "x" we need to use the quadratic formula (Attached), where a, b, and c, are the following:

a: 2 b: 3 c: 14

Using the quadratic formula, we get:

x1= 14/4 and X2 = -2

8 0
3 years ago
PLEASE HELP ASAP In this task, you will practice finding the area under a nonlinear function by using rectangles. You will use g
mrs_skeptik [129]

Answer:

a) 1280 u^{2}

b) 1320 u^{2}

c) \frac{4000}{3} u^{2}

Step-by-step explanation:

In order to solve this problem we must start by sketching the graph of the function. This will help us visualize the problem better. (See attached picture)

You can sketch the graph of the function by plotting as many points as you can from x=0 to x=20 or by finding the vertex form of the quadratic equation by completing the square. You can also do so by using a graphing device, you decide which method suits better for you.

A)

So we are interested in finding the area under the curve, so we divide it into 5 rectangles taking a right hand approximation. This is, the right upper corner of each rectangle will touch the graph. (see attached picture).

In order to figure the width of each rectangle we can use the following formula:

\Delta x=\frac{b-a}{n}

in this case a=0, b=20 and n=5 so we get:

\Delta x=\frac{20-0}{5}=\frac{20}{5}=4

so each rectangle must have a width of 4 units.

We can now calculate the hight of each rectangle. So we figure the y-value of each corner of the rectangles. We get the following heights:

h1=64

h2=96

h3=96

h4= 64

h5=0

so now we can use the following formula to find the area under the graph. Basically what the formula does is add the areas of the rectangles:

A=\sum^{n}_{i=1} f(x_{i}) \Delta x

which can be rewritten as:

A=\Delta x \sum^{n}_{i=1} f(x_{i})

So we go ahead and solve it:

A=(4)(64+96+96+64+0)

so:

A= 1280 u^{2}

B) The same procedure is used to solve part B, just that this time we divide the area in 10 rectangles.

In order to figure the width of each rectangle we can use the following formula:

\Delta x=\frac{b-a}{n}

in this case a=0, b=20 and n=10 so we get:

\Delta x=\frac{20-0}{10}=\frac{20}{10}=2

so each rectangle must have a width of 2 units.

We can now calculate the hight of each rectangle. So we figure the y-value of each corner of the rectangles. We get the following heights:

h1=36

h2=64

h3=84

h4= 96

h5=100

h6=96

h7=84

h8=64

h9=36

h10=0

so now we can use the following formula to find the area under the graph. Basically what the formula does is add the areas of the rectangles:

A=\sum^{n}_{i=1} f(x_{i}) \Delta x

which can be rewritten as:

A=\Delta x \sum^{n}_{i=1} f(x_{i})

So we go ahead and solve it:

A=(2)(36+64+84+96+100+96+84+64+36+0)

so:

A= 1320 u^{2}

c)

In order to find part c, we calculate the area by using limits, the limit will look like this:

\lim_{n \to \infty} \sum^{n}_{i=1} f(x^{*}_{i}) \Delta x

so we start by finding the change of x so we get:

\Delta x =\frac{b-a}{n}

\Delta x =\frac{20-0}{n}

\Delta x =\frac{20}{n}

next we find x^{*}_{i}

x^{*}_{i}=a+\Delta x i

so:

x^{*}_{i}=0+\frac{20}{n} i=\frac{20}{n} i

and we find f(x^{*}_{i})

f(x^{*}_{i})=f(\frac{20}{n} i)=-(\frac{20}{n} i)^{2}+20(\frac{20}{n} i)

cand we do some algebra to simplify it.

f(x^{*}_{i})=-\frac{400}{n^{2}}i^{2}+\frac{400}{n}i

we do some factorization:

f(x^{*}_{i})=-\frac{400}{n}(\frac{i^{2}}{n}-i)

and plug it into our formula:

\lim_{n \to \infty} \sum^{n}_{i=1}-\frac{400}{n}(\frac{i^{2}}{n}-i) (\frac{20}{n})

And simplify:

\lim_{n \to \infty} \sum^{n}_{i=1}-\frac{8000}{n^{2}}(\frac{i^{2}}{n}-i)

\lim_{n \to \infty} -\frac{8000}{n^{2}} \sum^{n}_{i=1}(\frac{i^{2}}{n}-i)

And now we use summation formulas:

\lim_{n \to \infty} -\frac{8000}{n^{2}} (\frac{n(n+1)(2n+1)}{6n}-\frac{n(n+1)}{2})

\lim_{n \to \infty} -\frac{8000}{n^{2}} (\frac{2n^{2}+3n+1}{6}-\frac{n^{2}}{2}-\frac{n}{2})

and simplify:

\lim_{n \to \infty} -\frac{8000}{n^{2}} (-\frac{n^{2}}{6}+\frac{1}{6})

\lim_{n \to \infty} \frac{4000}{3}+\frac{4000}{3n^{2}}

and solve the limit

\frac{4000}{3}u^{2}

4 0
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(-5x^6+7x+1)-(8x^2+x-10)<br><br> Algebra 1 subtracting polynomials please help
BaLLatris [955]

Answer:

-5x^6-8x^2+6x+11

Step-by-step explanation:

(-5x^6+7x+1)-(8x^2+x-10)

Remove parentheses

-5x^6+7x+1-(8x^2+x-10)

-(8x^2+x-10)=-8x^2-x+10

-5x^6+7x+1-8x^2-x+10

Simplify

-5x^6-8x^2+6x+11

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