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mr Goodwill [35]
3 years ago
7

W(n)=n^2-3n find w(4)

Mathematics
1 answer:
Zina [86]3 years ago
4 0

If w(n) = n² - 3n,

w(4) = 4² - 3(4)

= 16 - 12

= 4

w(4) = 4

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What is true about the solution of x^2/2x-6=9/6x-18?
pogonyaev

Answer:

  • x = ±√3, and they are actual solutions
  • x = 3, but it is an extraneous solution

Step-by-step explanation:

The method often recommended for solving an equation of this sort is to multiply by the product of the denominators, then solve the resulting polynomial equation. When you do that, you get ...

... x^2(6x -18) = (2x -6)(9)

... 6x^2(x -3) -18(x -3) = 0

...6(x -3)(x^2 -3) = 0

... x = 3, x = ±√3

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Alternatively, you can subtract the right side of the equation and collect terms to get ...

... x^2/(2(x -3)) - 9/(6(x -3)) = 0

... (1/2)(x^2 -3)/(x -3) = 0

Here, the solution will be values of x that make the numerator zero:

... x = ±√3

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So, the actual solutions are x = ±3, and x = 3 is an extraneous solution. The value x=3 is actually excluded from the domain of the original equation, because the equation is undefined at that point.

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<em>Comment on the graph</em>

For the graph, we have rewritten the equation so it is of the form f(x)=0. The graphing program is able to highlight zero crossings, so this is a convenient form. When the equation is multiplied as described above, the resulting cubic has an extra zero-crossing at x=3 (blue curve). This is the extraneous solution.

7 0
3 years ago
Read 2 more answers
Brian invests ?1900 into a savings account. The bank gives 3.5% compound interest for the first 2 years and 4.9% thereafter. How
Scorpion4ik [409]

let's check how much is it after 2 years firstly.


\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &1900\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases} \\\\\\ A=1900\left(1+\frac{0.035}{1}\right)^{1\cdot 2}\implies A=1900(1.035)^2\implies A=2035.3275


Brian invested the money for 6 years, so now let's check how much is that for the remaining 4 years.


\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &2035.3275\\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}


\bf A=2035.3275\left(1+\frac{0.049}{1}\right)^{1\cdot 4}\implies A=2035.3275(1.049)^4 \\\\\\ A\approx 2464.54\implies \boxed{\stackrel{\textit{rounded up }}{A=2465}}

4 0
3 years ago
I need help with math please
Dennis_Churaev [7]

Answer:

47/30

Step-by-step explanation:

2.35 * 2/3 =

= 2 35/100 * 2/3

= 2 7/20 * 2/3

= 47/20 * 2/3

= 94/60

= 47/30

5 0
3 years ago
Strontium-90 has a half-life of about 29 years. After 65 years, about how many grams of a 25-gram sample will remain?
konstantin123 [22]

Answer:

Ending Amount = Beginning Amount / 2 ^ n

where 'n' = number of half-lives n = 65 / 29 = 2.2413793103

Ending Amount = 25 g / 2^2.2413793103

Ending Amount = 25 / 4.7284892286

Ending Amount = 5.2871009727  g = 5.29 g (rounded)

Step-by-step explanation:

3 0
3 years ago
Fill in the blanks in the following proof, which shows that the sequence defined by the recurrence relation
lawyer [7]

Answer:

srry

Step-by-step explanation:

6 0
3 years ago
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