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2 years ago

SIMPLIFY THE FOLLOWING EXPRESSION

Mathematics

2 answers:

tatuchka [14]2 years ago

5 0

Answer:

-37m+6

Step-by-step explanation:

mkay so when working with simplifying any expressions like that, you wanna start with getting rid of your brackets. To do that, you simply take the number directly outside the bracket and you multiply it by everything inside. Let's take -9(m+2) first. You have to multiply everything inside the bracket by -9, so you have -9*m and -9*2, so you end up with -9m and -18. Now you do that to the second part of the equation (4*6 and 4*-7m) and you end up with 24 and -28m. Now you join all your terms together, and you have -9m-18+24-28m. Now all you do is add like terms (so -9m + -28m and -18+24) and you end up with a simplified expression of -37m+6. I hope this helped a little :)).

erik [133]2 years ago

4 0

Answer:

-37m+6

Step-by-step explanation:

-9(m+2)+4(6+7m)=Remove the parentheses.

-8-18+24-28m=Collect like terms and calculate.

-3m+6=solution

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3 years ago

How can I find the x-intercept in a graph if the equation is 2x+3y=6

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Step-by-step explanation:

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3 years ago

Use the drawing tool(s) to form the correct answer on the provided coordinate plane.

vlada-n [284]

Answer:

Step-by-step explanation:

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7 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 <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

Betsy drove from london to wolverhampton.it took her 2.5 hours at an average speed of 56 miles per hour.Mac drove the same journ

sergiy2304 [10]

Answer:

44.8 mph

Step-by-step explanation:

Betsy drove from london to wolverhampton.it took her 2.5 hours at an average speed of 56 miles per hour.Mac drove the same journy by motor bike.It took him 2 hours.Assuming that both took the same route,work out Mac's average speed in mph.

The above question is solved using Proportion method

2.5 hours = 56 mph

2 hours = x mph

Cross Multiply

2.5 hours × x mph = 2 hours × 56 mph

x mph = 2 hours × 56 mph/2.5 hours

x mph = 44.8 mph

Therefore, Mac's average speed = 44.8 mph

4 0

2 years ago