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3 years ago

You can tell that pi is an irrational because it has a what

1 answer:

3 0

I'll guess the answer is you can tell that pi is an irrational because it has a non-terminating yet non-repeating decimal representation.

Of course it's not clear how we tell this. We can't know for sure just by looking at the first trillion digits we've figured out whether it repeats or not. Someone told us it didn't, that's really how we know.

You might be interested in

If angle a has a measurement of 42 degrees and is complentary to angle b , what's the measure of angle b ?

USPshnik [31]

B is 90-42=48° since complementary angles add to 90°

5 0

3 years ago

at a point P on the parabola x^2=4ay a normal PK is drawn. From vertex O, a perpendicular OM is drawn to meet the normal at M. S

saveliy_v [14]

Answer: Its too long to write here, so I will just state what I did. I let P=(2ap,ap^2) and Q=(2aq,aq^2) But x-coordinates of P and Q differ by (2a) So P=(2ap,ap^2) BUT Q=(2ap - 2a, aq^2) So Q=(2a(p-1), aq^2) which means, 2aq = 2a(p-1) therefore, q=p-1 then I subbed that value of q in aq^2 so Q=(2a(p-1), a(p-1)^2) and P=(2ap,ap^2) Using these two values, I found the midpoint which was: M=( a(2p-1), [a(2p^2 - 2p + 1)]/2 ) then x = a(2p-1) rearranging to make p the subject p= (x+a)/2a

6 0

3 years ago

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 ...

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

_____

Comment on problem g

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

6 0

3 years ago

Which method would you choose to solve the following system?

gladu [14]

Answer: combining like terms

Step-by-step explanation:

3 0

3 years ago

Write an inequality for the graph

natima [27]

The open circle means the inequality will be greater than or equal to (≥) or less than or equal to (≤).
A closed circle means the inequality will be greater than (>) or less than (<)

An arrow pointing right to increasingly positive values means the inequality is getting greater (> or ≥)
A narrowing pointing left to increasingly negative values means the inequality is getting lesser (< or ≤)

So for this graph with an open circle and rightward pointing arrow, “x” will be some number on the number line greater than the first point of -38:
x > -38

6 0

3 years ago