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vodka [1.7K]
1 year ago
13

Segment AB falls on line 2x − 4y = 8. Segment CD falls on line 4x + 2y = 8. What is true about segments AB and CD? They are perp

endicular because they have the same slope of −2. They are perpendicular because they have slopes that are opposite reciprocals of −2 and one half. They are lines that lie exactly on top of one another because they have the same slope and the same y-intercept. They are lines that lie exactly on top of one another because they have the same slope and a different y-intercept
Mathematics
2 answers:
Makovka662 [10]1 year ago
8 0

Answer:

They are perpendicular because they have slopes that are opposite reciprocals of −2 and one half.

Step-by-step explanation:

Let's solve each equation for y and put it in the y = mx + b form. Then m is the slope, and we can tell if the lines are parallel or perpendicular or neither.

2x - 4y = 8

-4y = -2x + 8

<em>y = 1/2 x - 2; </em><em>m = 1/2</em>

4x + 2y = 8

2y = -4x + 8

<em>y = -2x + 4; </em><em>m = -2</em>

Now that both equations are in the slope-intercept form, we see that the slopes are 1/2 and -2.

The slopes are opposite reciprocals, -2 and 1/2, so the lines are perpendicular.

Answer: They are perpendicular because they have slopes that are opposite reciprocals of −2 and one half.

Alisiya [41]1 year ago
6 0

Answer:

They are perpendicular because they have slopes that are opposite reciprocals of −2 and one half.

Step-by-step explanation:

This is because x = -2 and half of -2 is 1

when we use CD line and x2 we find 8x+4y=16 when added to 2x -4y=8 would equal 10x+4y = 2 1/2 xy = 16

When we use for AB line we see they are perpendicular 2 1/2 x 2 = 5 -4y = 8 shows y to be -2 and the 1/2 line leaves -2 1/2 and x also is 2 1/2.

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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
2 years ago
The area of a square is stored in a double variable named area. write an expression whose value is length of the diagonal of the
lbvjy [14]

First of all, a bit of theory: since the area of a square is given by

A = s^2

where s is the length of the square. So, if we invert this function we have

s = \sqrt{A}.

Moreover, the diagonal of a square cuts the square in two isosceles right triangles, whose legs are the sides, so the diagonal is the hypothenuse and it can be found by

d = \sqrt{s^2+s^2} = \sqrt{2s^2} = s\sqrt{2}

So, the diagonal is the side length, multiplied by the square root of 2.

With that being said, your function could be something like this:

double diagonalFromArea(double area) {

double side = Math.sqrt(area);

double diagonal = side * Math.sqrt(2);

return diagonal;

}

3 0
1 year ago
A garden measuring 16 meters by 8 meters is going to have a walkway constructed all around the perimeter, increasing the total a
Semenov [28]

Answer:

The width of the pathway is:

  • <u>7 meters</u>.

Step-by-step explanation:

To identify the width of the pathway, you must remember the area formula of a rectangle:

  • Area of a rectangle = length * width.

From which the width can be cleared:

  • Width of a rectangle = area / length.

We know that the length of the terrain was not modified since the pathway is in the perimeter of the rectangle (16 m) and that the new area is 240 m^2, so we only have to replace the cleared formula:

  • Width of a rectangle = 240 m^2 / 16 m = 15 m.

The new width is equal to 15 meters, but since the question is not the total width but the width of the pathway, the width of the previously provided land must be subtracted from the value obtained.

  • <u>Pathway width = Total width - Garden width. </u>
  • <u>Pathway width = 15m - 8m = 7 meters.</u>
7 0
2 years ago
If a cow has a mass of 9×102 kilograms, and a blue whale has a mass of 1.8×105 kilograms, which of these statements is true?
lana66690 [7]

Answer:

The mass of the Blue whale is 200 times the mass of the cow

Step-by-step explanation:

Given

Mass of Cow = 9 * 10² kg

Mass of Blue Whale = 1.8 * 10⁵ kg

Required

Determine the relationship between both weights

Represent the mass of the cow with C and the mass of the whale with B

C = 9 * 10^2kg

C = 9 *100kg

C = 900kg

B = 1.8 * 10^5kg

B = 1.8 * 100000kg

B = 180000kg

Divide the bigger weight by the smaller weight

\frac{B}{C} = \frac{180000kg}{900kg}

\frac{B}{C} = \frac{180000}{900}

\frac{B}{C} = {200}{}

Multiply both sides by C

C * \frac{B}{C} = {200}{} * C

B = {200}{} * C

B = 200}C

<em>From the expression above, it can be concluded that the mass of the Blue whale is 200 times the mass of the cow</em>

7 0
1 year ago
Here is a scale drawing of a window frame that uses a scale of 1 cm to 6 inches.Create another scale drawing of the window frame
Tresset [83]

Answer:

Step-by-step explanation:

Below is the rectangle in the attachment.

Current scale:

1 cm : 6 inches

If the dimensions of the rectangle is:

Length = a cm

Width = b cm

Using the scale:

Length = a × 6 inches

Width = b × 6 inches

Using the same dimensions of the rectangle is:

Length = a cm

Width = b cm

Using the scale:

Length = a × 12 inches

Width = b × 12 inches

Note that there is an enlargement of the rectangle to form the new rectangle. The length and width of new rectangle drawn will be 2 × the length and width of the rectangle seen below.

8 0
1 year ago
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