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2 years ago

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Find the multiplier that should be used and the initial value, then write and exponential equation in the form f(x)=ab to the po

wer of x... the function passes through the points (1, 20) and (4, 2500)

1 answer:

astra-53 [7]2 years ago

4 0

Hdhdhdhdhdbdbdjhdddd

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Help with all please

seropon [69]

a) First, draw a graph. The x axis should be numbered: 0, 1, 2, 3, 4, 5, 6, 7, 8. Each number is the last digit of the year. For each year, there is one average score that has to be plotted. Number the y axis of the graph from 0-600, by increments of 50, or something around there. Now, with each year (each x value) place the dot as high as the average score. Repeat with all years. DO NOT draw a line to connect them. A scatter plot is a bunch of points that are not connected.

b) Use the form (y2-y1)/(x2-x1) where (x1,y1) (x2,y2) are any two points from 2001-2006. It does not matter which points, pick any, and assign each of the numbers x1,y1 and x2,y2. Plug it in to the equation, and simplify. This is your slope, also called m. Now plug m into the equation y-y1=m(x-x1) where y1 and x1 are the x and y coordinates of any point from 2001 to 2006. x and y dont have values themselves, they stay there. Now distribute the m into the parenthesis, add y1 to both sides (to cancel it out on the left), and you should be left with the same equation, but now in y=mx+b form. B is where the line crosses the y axis, so put a point on the vertical axis where b is. M is your slope, so every time you go to the right one number, go up M numbers. Draw another point. Repeat this until you can connect these new dots into a line. This line should be on the same graph as your other points, but might not touch all of the scatter plot points. That's still okay, just leave it as is.

c) Find the point where x=6 (meaning the year is 2006) on your line. Find the average test score for that year by seeing where the point lines up along the y axis (draw a straight line from the point to the left to see, if you have to). Take that number and add 1m (the m is the number you used in step B) to it. This predicts what the average score might be after 1 (that's why the 1 is there) year. This is a predicted value, and might not be perfectly correct!! Write/record how close this new, predicted, score is to the actual number, 515, which can be found on the scatter plot.

d) repeat step C exactly as before (still use the year 2006) , but add 4m instead of 1m, to get the predicted score 4 years after 2006 (2010), instead of 1 year after (2007).

I hope I have been of some help! Best of luck!!! :)

3 0

3 years ago

Describe the transformation that maps the pre image A to the image A'

Leya [2.2K]

Answer:

B) translated 6 units to the right then translated 9 units up

Step-by-step explanation:

Just use the graph to follow each translation until you get thee right answer, or you can count how many spaces are from the pre-image to the image. Just make sure to always use the same point.

4 0

3 years ago

Read 2 more answers

The radius,r , of a sphere is 4 . 3m Calculate the sphere's volume,v . Use the value 3.14 for sqaure, and round your answer to t

kotykmax [81]

Answer:

V(s) = 332,9 m³

Step-by-step explanation:

The volume of a sphere is:

V(s) = (4/3)*π*r³

if we know r = 4,3 m then

V(s) = (4/3)*3,14*(4,3)³

V(s) = 332,8693 m³

Rounding the answer to the nearest tenth

V(s) = 332,9 m³

4 0

2 years ago

Order the rational numbers from least to greatest<br> 0.42, -1/2, 11/20, and -0.51

umka21 [38]

-0.51, -1/2, 0.42, 11/20

7 0

3 years ago

Read 2 more answers

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

3 0

3 years ago