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3 years ago

Complete the square to determine the minimum or maximum value of the function defined by the expression. −x2 + 18x − 99

2 answers:

6 0

Given the expression:
-x^2+18x-99
to solve by completing squares we proceed as follows:
-x^2+18x-99=0
this can be written as:
-x^2+18x=99
x^2-18x=-99.......i
but
c=(-b/2)²
Hence:
c=(-(-18)/2)²=81
adding 81 in both sides of i we get:
x^2-18x+81=-99+81
factorizing the quadratic we obtain:
(x-9)(x-9)=-18
thus
(x-9)²+18=0
the above takes the vertex form of :
y=(x-k)²+h
where (k,x) is the vertex:
the vertex of our expression is:
(9,18)
hence the maximum point is at (9,18)
NOTE: The vertex gives the maximum point because, from the expression we see that the coefficient of the term that has the highest degree is a negative, and since our polynomial is a quadratic expression then our graph will face down, and this will make the vertex the maximum point.

Serggg [28]3 years ago

4 0

Answer:

answer is C hope you have a good day

Step-by-step explanation:

it is Maximum -18

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Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for

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Answer:

a) The recurrence formula is $P_n = \frac{109}{100}P_{n-1}$ .

b) The general formula for the population of Tacoma is

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$ . For the population in 2001 we will use $P_1$ , for the population in 2002 we will use $P_2$ , and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$ .

In 2002, we will have the population of 2001, $P_1$ , plus the 9% of $P_1$ . This is

$P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1$ .

So, it is not difficult to notice that the general recurrence is

$P_n = \frac{109}{100}P_{n-1}$ .

b) In the previous formula we only need to substitute the expression for $P_{n-1}$ :

$P_{n-1} = \frac{109}{100}P_{n-2}$ .

Then,

$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$ .

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$ .

But we can do the same operation n times, so

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) Recall the notation we have used:

$P_{0}$ for 2000, $P_{1}$ for 2001, $P_{2}$ for 2002, and so on. Then, 2016 is $P_{16}$ . So, in order to obtain the approximate population of Tacoma in 2016 is

$P_{16} = \left(\frac{109}{100}\right)^{16}P_{0} = (1.09)^{16}P_0 = 3.97\cdot 200000 \approx 794062$

d) In this case we want to know when $P_n>400000$ , which is equivalent to

$(1.09)^{n}P_0>400000$ .

Substituting the value of $P_0$ , we get

$(1.09)^{n}200000>400000$ .

Simplifying the expression:

$(1.09)^{n}>2$ .

So, we need to find the value of $n$ such that the above inequality holds.

The easiest way to do this is take logarithm in both hands. Then,

$n\ln(1.09)>\ln 2$ .

So, $n>\frac{\ln 2}{\ln(1.09)} = 8.04323172693$ .

So, the population of Tacoma should exceed the 400 000 by the year 2009.

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Answer:

Each monthly payment is $ 49.05, and in total Tom paid $ 648.60.

Step-by-step explanation:

Given that Tom buys a DVD player on 12 monthly payments, whose cash price is $ 600, and he pays a 10% deposit and the store charges 9% interest, to determine how much is each payment and the total amount Tom pays for the DVD player the following calculations must be performed:

600 - 600 x 0.1 = 540

540 x 1.09 = 588.6

588.6 / 12 = 49.05

Thus, each monthly payment is $ 49.05, and in total Tom paid $ 648.60.

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