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1 year ago

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On Monday, Rachel bought company stock at a price of $39.60 per share. On Tuesday, the share price changed by -$2.35. On Wednesd

ay, the share price changed by +$1.45. What is the share price at the end of Wednesday?

1 answer:

AfilCa [17]1 year ago

5 0

39.60 - 2.35 + 1.45 = $38.70

Answer: The price at the end of Wednesday is $38.70

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Math 9:The floor of conference hall can be covered completely with tiles. Its length is 36 ft longer than its width. The area of

valentina_108 [34]

What is the question?
I'm assuming it is to find the length and width.
+_= plus or minus
(X+36)
____________
| |
(X) | |
|____________|

X^2+36X-2040<0
X<-36+_(36^2-4*-2040)^(1/2)
-----------------------------------
2
X<-18+_2((591)^(1/2))
This is probably not what you wanted, sorry

8 0

2 years ago

Maya can run 18 miles in 3 hours, and she can bike 18 miles in 2 hours.

Aloiza [94]

Answer:

36

Step-by-step explanation:

if it is two hours then half of it is 9

so 18 + 9 = 27 which is how many miles she biked

6 is the number of miles she ran

so u times it by 1.5

27 + 9 = 36

4 0

1 year ago

Read 2 more answers

Kyle asks his friend Jane to guess his age and his grandmother’s age. Kyle says his grandmother is not more than 80 years old. H

melamori03 [73]

#1 is incorrect. The first equation says that his grandmother is not more than 80 years old. If you confine yourself to whole numbers, then she could be 80 but not more. That inequality should be written as g < 81 or if you are a purist, g≤ 80.

The second equation is correct. At the very most her age is equal to 3k - 3 so at most it would be written as g = 3k - 3.

Since it is not that much, you would write it as g ≤ 3k - 3

6 0

2 years ago

Read 2 more answers

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

Denise is constructing A square in which two of its vertices are points M and N. She has already used her straightedge and compa

yan [13]

Denise is constructing A square.

Note: A square has all sides equal.

We already given two vertices M and N of the square.

And another edge of the square is made by from N.

Because a square has all sides of equal length, the side NO should also be equal to MN side of the square.

Therefore, <em>Denise need to place the point of the compass on point N and draw an arc that intersects N O, using MN as the width for the opening of the compass. That would make the NO equals MN.</em>

Therefore, correct option is :

D) place the point of the compass on point N and draw an arc that intersects N O, using MN as the width for the opening of the compass.

6 0

2 years ago

Read 2 more answers