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4 years ago

Hey everyone hopefully your doing great just want to ask a quick question and hopefully you guys answer!!!

Mathematics

1 answer:

MrRa [10]4 years ago

8 0

Answer:

m<XDQ = 41*

m<UXD = 138*

Step-by-step explanation:

for m<XDQ... 90-49= 41

for m<UXD... 180-41=139

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Find 0.1 more than a number and 0.1 less than a number. Find 0.01 more than a number and 0.01 less than a number. Find 0.001 mor

viktelen [127]

Answer:

Okay so our number in this example is going to be 23 because it is my favorite.

So you start of by saying okay

23+0.1

and that would equal 23.1

and then you would think

23-0.1

and that would equal 22.9

and then you would do the same thing just move a place

so 23+0.001

and that would equal 23.001

and then you subtract it

so 23-0.001

and that would equal 22.999

hope this helped :)

6 0

1 year ago

Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

3 0

3 years ago

Find the measures of the numbered angles in the figure, shown to the right. M<1= M<2= M<3= M<4

const2013 [10]

Answer:

$\angle 1 = 101$

$\angle 2 = 101$

$\angle 3 = 101$

$\angle 4 = 101$

Step-by-step explanation:

Given

The attachment

Require

Solve for $\angle 1, \angle 2, \angle 3, \angle 4$

We start by solving for $\angle 1$

$\angle 1 + 79 = 180$ ---- Angle on a straight line

$\angle 1 + 79-79 = 180 - 79$ --- Subtract 79 from both sides

$\angle 1 = 101$

$\angle 1 = \angle 2 = \angle 3$ --- Corresponding angles;

So, we have:

$\angle 2 = 101$

$\angle 3 = 101$

$\angle 1 = \angle 4$ --- vertically opposite angles

So, we have:

$\angle 4 = 101$

4 0

3 years ago

5. Find the value(s) of x so that the line containing the points (2x + 3, x + 2) and (0, 2) is

Dvinal [7]

Answer:

x = -2 or -9

Step-by-step explanation:

You want the values of x such that the line defined by the two points (2x+3, x+2) and (0, 2) is perpendicular to the line defined by the two points (x+2, -3-3x) and (8, -1).

<h3>Slope</h3>

The slope of a line is given by the slope formula:

m = (y2 -y1)/(x2 -x1)

Using the formula, the slopes of the two lines are ...

m1 = (2 -(x+2))/(0 -(2x+3)) = (-x)/(-2x-3) = x/(2x +3)

and

m2 = (-1 -(-3-3x))/(8 -(x+2)) = (2+3x)/(6 -x)

<h3>Perpendicular lines</h3>

The slopes of perpendicular lines have product of -1:

$\dfrac{x}{2x+3}\cdot\dfrac{2+3x}{6-x}=-1\\\\x(3x+2)=(2x+3)(x-6)\qquad\text{multiply by $(2x+3)(6-x)$}\\\\3x^2+2x=2x^2-9x-18\qquad\text{eliminate parentheses}\\\\x^2+11x+18=0\qquad\text{put in standard form}\\\\(x+2)(x+9)=0\qquad\text{factor}$

<h3>Solutions</h3>

The values of x that satisfy this equation are x = -2 and x = -9. The attached graphs show the lines for each of these cases.

4 0

1 year ago

14 and 15 pls and I need work too thax

anastassius [24]

14.
number of grams in box / number of grams in serving = 512/56

512/56

Use long division to find your answer.

___9__ remainder: 8 = 8/56
56 I 512
-504
-------
8

Final answer: 9 8/56 servings in a box

6 0

3 years ago