Ask question

Ask question

All categories

2 years ago

15

Alexa buys 27 stamps, each stamp is of 1 square unit of area. She exchanges them for 8 sugar cubes, whose total volume is numeri

cally (not dimensionally) equal to the total area of Alexa?s stamps. Which of the following represents the length of a cube of sugar

1 answer:

Vikentia [17]2 years ago

8 0

The answer is 8 sugar cubes

You might be interested in

Look at the figure. Which of the following segments is parallel to AB?

marusya05 [52]

I think it’s c and h

5 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

1 year ago

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find

fenix001 [56]

Answer:

y2 = (6x + 7)/36 + (Dx + E)e^x

Step-by-step explanation:

The method of reduction of order is applicable for second-order differential equations.

For a known solution y1 of a 2nd order differential equation, this method assumes a second solution in the form Uy1 which satisfies the said differential equation. It then assumes a reduced order for U'' (w' = U'').

The differential equation becomes easy to solve, and all that is left are integration and substitutions.

Check attachments for the solution to this problem.

6 0

1 year ago

Here are the 30 best lifetime baseball batting averages of all time, arranged in order from lowest to highest: 0.329, 0.330, 0.3

dsp73

When putting data into a class for this case, 0.350 - 0.359, they should be within the class boundaries. The class boundaries are 0.3495 and 0.3595. So, the data that will go into that class are 0.356 and 0.358. The record 0.349 is not included since it is below 0.3495. Therefore, there are only two values in the class.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

4 0

1 year ago

Read 2 more answers

Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant

pychu [463]

Answer:

A=152

K= -Ln(0.5)/14

Step-by-step explanation:

You can obtain two equations with the given information:

T(14 minutes) = 114◦C

T(28 minutes)=152◦C

Therefore, you have to replace t=14, T=114 and t=28, T=152 in the given equation:

$114=190-Ae^{-14k} (I) \\152=190-Ae^{-28k}(II)$

Applying the following property of exponentials numbers in (II):

$e^{a}.e^{b}=e^{a+b}$

Therefore $e^{-28k}$ can be written as $e^{-14k}.e^{-14k}$

$152=190-Ae^{-14k}.e^{14k}$

Replacing (I) in the previous equation:

$152=190-76e^{-14k}$

Solving for k:

Subtracting 190 both sides, dividing by -76:

$0.5=e^{-14k}$

Applying the base e logarithm both sides:

Ln(0.5)= -14k

Dividing by -14:

k= -Ln(0.5)/14

Replacing k in (I) and solving for A:

$Ae^{-14(-Ln(0.5)/14)}=76\\Ae^{Ln(0.5)} =76\\A(0.5)=76$

Dividing by 0.5

A=152

7 0

1 year ago