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2 years ago

5

Please help me out with this because I have this due today and I’m trying to finish either way I wanna make sure if it’s right o

r either wrong... thank you

1 answer:

Nostrana [21]2 years ago

7 0

you have it all correct so far, except for part b. it's plan b instead of plan a i believe

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Answer the following questions CORRECTLY I will know if this is wrong. I WILL REPORT ANY INCORRECT ANSWERS!

Anna71 [15]

Answer by JKismyhusbandbae: B) –10.6

Work/Explanation: Since the sequence slowly gets smaller, it is likely that each term is something added to (subtracted from) the previous term. To get from –7.9 to –8.8, it appears that the first has had –0.9 added to it. Continue adding –0.9 to get that the fourth term is –10.6.

8 0

2 years ago

21. Randy owns a computer store. In 1990, he sold 150 monitors. In 2000, he sold 900

lutik1710 [3]

Answer:

$y=75x+150$

Step-by-step explanation:

Let $y$ represent the number of monitors sold.

Given:

$x$ is the number of years since 1990.

So, for the year 1990, $x=0$ .

For the year 2000, 10 years passed. So, $x=10$ .

Now, monitors sold in 1990 are 150. So, at $x=0,y=150$

Monitors sold in 2000 are 900. So, at $x=10,y=900$

Thus, the two points are $(0,150)$ and $(10,900)$ .

The slope of a line with points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is given as:

Slope, $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

For the points $(0,150)$ and $(10,900)$ , the slope is given as:

Slope, $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\ m=\frac{900-150}{10-0}\\ m=75$

Now, the y-intercept is the point where $x=0$ . So, the point $(0,150)$ has $x$ value as 0.

So, the y-intercept is 150.

Equation of a line in slope-intercept form is given as:

$y=mx+b$

Where, $m$ is the slope and $b$ is the y intercept.

Here, $m=75$ and $b=150$ .

Therefore, the equation that represents the above data is given as:

$y=75x+150$

7 0

3 years ago

What is the equation for this graph?<br><br> A. y=2x + 1<br> B. y=1/2x + 1<br> C. y=1/2x<br> D. y=2x

MrRa [10]

B is the answer :)))

8 0

3 years ago

Find the slope of the line that contains the points:<br> (-4,3) and (4,9)

Stolb23 [73]

I believe the slope is 3/4

9-3 = 6
4 - (-4) = 8

Simplify to get 3/4

3 0

3 years ago

Read 2 more answers

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(C) $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

4 0

3 years ago