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mr_godi [17]
3 years ago
10

Which is an irrational number

Mathematics
1 answer:
astraxan [27]3 years ago
8 0

Answer:

C

Step-by-step explanation:

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How to solve for 1000(x-2)=75000
Andrews [41]
So, first you divide 75000 by 1000. Then you add 2 to each side. Then that's the answer, x=77
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3 years ago
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..Urgent.. 600 students attend Ridgewood Junior High School. 10% of students bring their lunch to school everyday. How many stud
zysi [14]

Answer:

Number of student bring their lunch = 60

Step-by-step explanation:

Given:

Total number of student = 600

Student bring their lunch = 10%

Find:

Number of student bring their lunch

Computation:

Number of student bring their lunch = Total number of student x Student bring their lunch

Number of student bring their lunch = 600 x 10%

Number of student bring their lunch = 60

3 0
3 years ago
The following triangles are similar. Find side AB.
tankabanditka [31]

Answer:

4

Step-by-step explanation:

Set up ratios of corresponding sides.  x+2 is to 6 as 2 is to 3:

\frac{x+2}{6}=\frac{2}{3}

Now cross multiply to get

3(x + 2) = 12 and

3x + 6 = 12 so

3x = 6 and

x = 2.  That means that side AB, x + 2, is 2 + 2 which is 4

4 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
PLEASEEE HELPPP<br><br> Solve for x:
m_a_m_a [10]

Second option, just do Pemdas backwards.

5 0
3 years ago
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