Check all of the ordered pairs that satisfy the equation below.<br> y= 6x

diketahui luas permukaan kubus 864 cm2. perbandingan rusuk kubus dan p balok 3 : 4 perbandingan p l dan tinggi 4:3:2 luas permuk

Kryger [21]

S2= 864/6
s = = 12 cm
p = 4/3 x 12 = 16 cm
t = 2/4 x 16 = 8 cm
l= 3/4x 16 = 12 cm

lp= 2 (pl+pt+lt)
= 2 (16 x 12 + 16 x 8 + 12 x8)
= 2 (192+128+96) = 832cm²

3 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 ...

a)

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

__

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 ...

d)

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

__

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

f)

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

_____

<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

3 years ago

7. The lifetime of a certain brand of tires is approximately normally distributed, with a

Aleonysh [2.5K]

Answer:

0.0025551

Step-by-step explanation:

Given that:

Mean (m) = 40000

Standard deviation (s) = 2500

P(x < 33,000) :

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (33000 - 40000) / 2500 = - 2.8

p(Z < -2.8) = 0.0025551 ( Z probability calculator)

Probability of Tyre wagering out before mileage limit is reached = 0.0025551

7 0

2 years ago

Why am i so pretty and funny ? :/

BabaBlast [244]

Answer:

Because you are amazing

7 0

3 years ago

Read 2 more answers

Find a formula for the inverse of the function. f(x)=1-2/x^3.

Elena-2011 [213]

f(x) = 1 - ²/ₓ₃
y = 1 - ²/ₓ₃
y = 1 - ²/ₓ₃
y - 1 = ⁻²/ₓ₃
x - 1 = -2/y³
y³(x - 1) = -2
y³ = ⁻²/ₓ₋₁
y = ∛⁻²/ₓ₋₁
y = -∛(2x² - 4x + 2)/x - 1
f⁻¹(x) = -∛(2x² - 4x + 2)/x - 1

4 0

3 years ago

Check all of the ordered pairs that satisfy the equation below. y= 6x