What is the answer to 4x + 27

Please me<br><br>please me noww

vesna_86 [32]

Step-by-step explanation:

5. answer ➡D

6 . answer ➡A

8 . answer ➡C

3 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 ...

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

Hey please help me please help please

Airida [17]

1.

a) metres to centimetres :

multiply length by 100

b) metres to millimetres:

multiply length by 1000

c) kilograms to grams:

multiply the mass value by 1000

d) litres to millilitres :

multiply volume by 1000

2.

a) 3 m = 3× 100 = 300 cm

b) 28 cm = 28 × 10 = 280 mm

c) 2.4 km = 2.4 × 1000

= 24 × 10^-1 × 10^3

= 24 × 10^2 =2400 m

d) 485 mm =485 / 10

= 485 / 10 ^1

= 485 × 10 ^-1

= 48.5 cm

e) 35 cm = 35 / 100

= 35 /10^2

= 35 × 10 ^ -2

= 0.35 m

f) 2.4 m = 2.4 / 1000

= 24 × 10 ^-1 / 10^3

= 24 × 10^-1 × 10 ^-3

= 24 × 10 ^ -4

= 0.0024 km

g) 2495 mm = 2495 /1000

= 2495 /10^ 3

= 2495 × 10 ^-3

=2.495 m

4 0

3 years ago

I need this please help me

kherson [118]

Answer:

I attached a picture of the completed picture below.

The values given to you in the picture are shown in pink

Use those values to fill in the rest, which are shown in blue.

6 0

3 years ago

Solve the following equation for X. Round your answer to four decimal places.

strojnjashka [21]

ANSWER

x = 1.2226

EXPLANATION

To solve this equation we have to apply the property of the logarithm of the base,

$\log _bb=1$

Thus, we can apply the natural logarithm - whose base is e, to both sides of the equation,

$\ln e^{4x}=\ln 133$

Now we apply the property of the logarithm of a power,

$\log a^b=b\log a$

In our equation,

$\begin{gathered} 4x\ln e^{}=\ln 133 \\ 4x^{}=\ln 133 \end{gathered}$

Then divide both sides by 4 and solve,

$\begin{gathered} \frac{4x^{}}{4}=\frac{\ln 133}{4} \\ x\approx1.2226 \end{gathered}$

The solution to this equation is x = 1.2226, rounded to four decimal places.

4 0

1 year ago