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3 years ago

How to solve -5x=9-2x

Mathematics

1 answer:

galina1969 [7]3 years ago

8 0

Answer:

The answer would be 3

Step-by-step explanation:

Add the -2x to the -5x because they are like terms

3x=9

Divide 9 by 3 to isolate the variable

x=3

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1 1/4 miles equals how many feet

Artist 52 [7]

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3 years ago

Describe the transformation of f(x) to g(x)

gulaghasi [49]

Answer:

D. f(x) is shifted up 1 unit to g(x).

Step-by-step explanation:

Just let's look at the y-intercept of each function.

f(0) = 0

g(0) = 1

Also remember that a vertical shift of N units is written as:

g(x) = f(x) + N

where if N is positive, the shift is upwards, and if N is negative, the shift is downwards.

then we can write, for the particular value x = 0

g(0) = f(0) + N

replacing the values

1 = 0 + N

1 = N

Then N is positive, so we have a shift up of one unit.

The correct option is:

D. f(x) is shifted up 1 unit to g(x).

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2 years ago

Please help me i will give brainliest *image

Alika [10]

7.5 Devide 30 by 4 and it is 7.5 that is how much she saved

8 0

3 years ago

Read 2 more answers

After 5 tests Dara's average score was 88 what score must she average on the next two tests to have a seven-test average of 90?

Tomtit [17]

After 5 tests, her average was 88. To have a seven-test average of 90, the total must be:

$90 \times 7 = 630$

Her average after 5 tests was 88. Her total score on the 5 tests is:

$88 \times 5 = 440$

Subtract:

$630-440 = 190$

Find the average by dividing by 2:

$190 \div 2 = 95$

Her average must be 95

3 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