All categories

3 years ago

Find the repeating decimal between -2 1/3 and -2 1/5.

1 answer:

6 0

let's first off, make the mixed fractions to improper fractions and then <u>multiply one by the other's denominator</u>, that way both fractions have the same denominator, so let's start,

$\bf \stackrel{mixed}{2\frac{1}{3}}\implies \cfrac{2\cdot 3+1}{3}\implies \stackrel{improper}{\cfrac{7}{3}}~\hfill \stackrel{mixed}{2\frac{1}{5}}\implies \cfrac{2\cdot 5+1}{5}\implies \stackrel{improper}{\cfrac{11}{5}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{multiplying one by the other's denominator}}{\cfrac{7}{3}\cdot \cfrac{5}{5}\implies \boxed{\cfrac{35}{15}}\qquad \qquad \cfrac{11}{5}\cdot \cfrac{3}{3}\implies \boxed{\cfrac{33}{15}}} \\\\[-0.35em] ~\dotfill$

$\bf \boxed{-\cfrac{35}{15}}\rule[0.35em]{9em}{0.25pt}~~\underset{\underset{\textit{\large -2.66666666...}}{\uparrow} }{-\cfrac{34}{15}}~~\rule[0.35em]{9em}{0.25pt}\boxed{-\cfrac{33}{15}}$

You might be interested in

The table shows the cost of downloading songs.

Sonbull [250]

Answer:

Step-by-step explanation:

rate of change=cost per song

so the first option

3 0

3 years ago

Given isosceles right triangle CDE with m∠D = 90, sin C = sin D

Butoxors [25]

Answer:True

Step-by-step explanation:

Given

$\angle D=90^{\circ}$

For isosceles triangle , CD=DE (x=y)

from figure,

$\sin C=\dfrac{y}{H}\\\\\sin D=\dfrac{x}{H}\\\\\text{CD=DE}\\\\\therefore \sin C=\sin D=\dfrac{x}{H}=\dfrac{y}{H}$

Thus, the given statement is true

7 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 ...

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