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1 year ago

7

Assume each tick mark represents 1 cm. Calculate the total displacement from 0 if an object moves 3 cm to the left, then 7 cm to

the right, and then 6 cm to the left

2 answers:

tatyana61 [14]1 year ago

6 0

Answer:

Calculate the final displacement from 0 if an object first moves 3 cm to the left and then 5 cm to the right.

The final displacement is 2 cm to the RIGHT .

Step-by-step explanation:

Arte-miy333 [17]1 year ago

4 0

Moving 3 cm to the left would be "-3", then 7 to the right makes it "4" and 6 to the left makes it "-2", so it would be 2 centimeters behind 0, the total displacement is "2" (negative doesn't apply because you can't be -2 cm away from something)

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2 years 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

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So we expect a general solution of the form

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

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

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 $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
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3 0

2 years ago

Triangle ABE is similar to triangle ACD. AED and ABC are straight lines. EB and DC are parallel. AE = 5 cm, BC = 4.5 cm, BE = 4

lianna [129]

Answer:

y = 16x/65

Step-by-step explanation:

Given:

Triangle ABE is similar to triangle ACD. AED and ABC are straight lines

EB and DC are parallel

The area of quadrilateral BCDE = xcm²

The area of triangle ABE = ycm²

Find attached the diagram from the above information.

In similar triangles, the ratio of their corresponding angles are equal.

Also, the ratio of the area of the two triangles = square of ratio of the corresponding sides of the two triangles.

Area ∆ACD/area of ∆ABE = (DC/EB)²

Area ∆ACD/area of ∆ABE = [(area of quadrilateral BCDE +

area of ∆ABE)]/(area of ∆ABE)

(x+y)/y = (DC/EB)²

(x+y)/y = (9/4)²

x+y = (81/16)y

x = (81/16)y - y

x = (81y - 16y)/16

x = 65y/16

Making y subject of formula

16x = 65y

y = 16x/65

An expression for y in terms of x:

y = 16x/65

3 0

1 year ago