All categories

3 years ago

What are some other factors to consider when contemplating a career?

Mathematics

1 answer:

Liono4ka [1.6K]3 years ago

5 0

Answer:

How feasible it is to achieve

What level of schooling you need

Health benefits

Amount you are paid

The overall work ethic

Step-by-step explanation:

You might be interested in

What is 2 ^ 4 in standard form?

spin [16.1K]

6x=10^= (*=0) is the answer

3 0

3 years ago

Caroline has 10 marbles. One half of them are blue. How many marbles are blue?

Viktor [21]

Answer: 5

One half of 10 is 10/2 or 1/2(10)
10/2 = 5
1/2(10) = 5

6 0

3 years ago

Read 2 more answers

Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

Using these formulas on problem (d), we get ...

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

And for problem (f), we get ...

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0

2 years ago

Make a table of values for each eqution<br> y = |x| + 1

Alisiya [41]

No x-intercept/Zero
y=1

Hope I can help you!

4 0

3 years ago

Someone do this got me rq

skelet666 [1.2K]

Answer:

The last listed functional expression:

$\left \{ {x+1\,\,\,\,{x\geq 2} \atop {x+2 \,\,\,\,x$

Step-by-step explanation:

It is important to notice that the two linear expressions that render such graph are parallel lines (same slope), and that the one valid for the left part of the domain, crosses the y-axis at the point (0,2), that is y = 2 when x = 0. On the other hand, if you prolong the line that describes the right hand side of the domain, that line will cross the y axis at a lower position than the previous one (0,1), that is y=1 when x = 0. This info gives us what the y-intercepts of the equations should be (the constant number that adds to the term in x in the equations: in the left section of the graph, the equation should have "x+2", while for the right section of the graph, the equation should have x+1.

It is also important to understand that the "solid" dot that is located in the region where the domain changes, (x=2) belongs to the domain on the right hand side of the graph, So, we are looking for a function definition that contains $x+1$ for the function, for the domain: $x\geq 2$ .

Such definition is the one given last (bottom right) in your answer options.

$\left \{ {x+1\,\,\,\,{x\geq 2} \atop {x+2 \,\,\,\,x$

7 0

3 years ago