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3 years ago

How does a slope of 9/2 compare to 3/2?

Mathematics

1 answer:

Softa [21]3 years ago

4 0

Answer: 3

Step-by-step explanation

9 divided by 2 =4.5

3 divided by 2 = 1.5

so then we do 4.5 divided by 1.5=3 so the answer is 3

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N+7< -3 Solve this inequality

Mice21 [21]

$n+7\ \textless \ -3 \\ \\ n \ \textless \ -3 - 7 \\ \\ n \ \textless \ -10 \\ \\$

The final result is: n < -10.

7 0

3 years ago

Read 2 more answers

5. What is the center of (0.1, 0.3, 0.02, 0.09, 0.4, 0.05, 0.26}? 0.2 0.3 0.1 0.4

Dima020 [189]

Answer:

0.09

Step-by-step explanation:

The data set is:

0.1, 0.3, 0.02, 0.09, 0.4, 0.05, 0.26

The number in the middle is 0.09. Hope this helps!

3 0

3 years ago

Can someone please help meee

Mariana [72]

Answer:

Step-by-step explanation:

Ok so you are given the values of the slope-intercept form with m being the slope and b being the y-intercept. So since b is equal to -1 you want to plot a point at (0, -1) since that is the y-intercept (when x = 0). The next thing you want to do is look at the slope, which is essentially saying each time x increases by 5 the y-value decreases by 4 or in other words rise/run which is negative which is why you're going down. So from the point (0, -1) go forward 5 units and go down 4 units which should lead you to (5, -5) and the third point you can plot is by going backwards instead of forwards. So instead of every time x increases by 5 y decreases by 4 you're going to do the inverse. Every time x decreases by 5, y is going to increase by 4. So by doing this from the y-intercept (0, -1) you should go backwards 5 units and up 4 units which should lead you to (-5, 3). And then now just draw a line that goes through all those three points. Hope that helps :)

6 0

1 year ago

Which is more 45g or 45ml?

MatroZZZ [7]

For water 1 gram = 1 ml.

This means 45 grams are equal to 45 ml's.

Neither one is greater than the other one as they are equal.

The problem doesn't state what is being measured, so the answer could be different depending on the density of the product being measured.

8 0

3 years ago

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 f 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