Confused with a question. Will upvote!

1 answer:

5 0

The first 3 months are 250
6 months are 500
9 months are 1000
12 months are 2000
15 months are 4000
18 months are 8000
21 months are 16000
24 months are 32,000 which is two years

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Write as an expression: the total of n and 16

Mekhanik [1.2K]

Total is just add
n+16 is total of 16 and n

4 0

3 years ago

What are the roots of the polynomial equation x^3-5x+5=2x^2-5? Use a graphing calculator and a system of equations. Round nonint

leonid [27]

The right answer is c. –2.24, 2, 2.24

This question needs to be solved in two ways. First, using a graphing calculator. Next, using a system of equations.

1. Using a graphing calculator.

We have the following polynomial equation:

$x^3-5x+5=2x^2-5$

By ordering this equation we have:

$x^3-2x^2-5x+10=0$

So, we can say that this equation comes from a function given by:

$f(x)=x^3-2x^2-5x+10$

Thus, by plotting this function, we have that the graph of this function is indicated in Figure 1. By zooming, we can see, in Figure 2, that the roots of the polynomial equation are the x-intercepts of $f(x)$ which are:

$x_{1}=-2.236 \\ \\ x_{2}=2 \\ \\ x_{3}=2.236$

Finally, rounding noninteger roots to the nearest hundredth we have:

$\boxed{Root_{1}=-2.24} \\ \\ \boxed{Root_{2}=2} \\ \\ \boxed{Root_{3}=2.24}$

2. Using a system of equations.

The ordered equation is:

$x^3-2x^2-5x+10=0$

By arranging to factor out we have:

$x^3-5x-2x^2+10=0$

Then, by factoring:

$x(x^2-5)-2(x^2-5)=0$

Term $(x^2-5)$ is a common factor, thus:

$(x-2)(x^2-5)=0 \\ \\ (x-2)(x-\sqrt{5})(x+\sqrt{5})=0 \\ \\ Finally: \\ \\ \boxed{Root_{1}=-\sqrt{5}=-2.24} \\ \\ \boxed{Root_{2}=2} \\ \\ \boxed{Root_{3}=\sqrt{5}=2.24}$