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3 years ago

find the factors of each number. Write the common factor ( CF ) and then state the greatest common factor ( GCF) .

Mathematics

1 answer:

Reptile [31]3 years ago

6 0

Given:

Two numbers are 12 and 21

To find:

The factors of 12 and 21, then find the common factor and the greatest common factor.

Solution:

Two numbers are 12 and 21. The prime factors of these two numbers are

$12=2\times 2\times 3$

$21=3\times 7$

From the above factorization, it is clear that the factor 3 is common in both. So,

Common factor (CF) = 3

Only 3 is common in factorization of both. So,

Greatest common factor (GCF) = 3

Therefore, the common factor is 3 and the greatest common factor is also 3.

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Write the equation of the line through the indicated point with the indicated slope. write the final answer in the form y equals

Assoli18 [71]

y = - 4x + 13

slope = -4

y-intercept = 13

8 0

3 years ago

Given the Arithmetic series A 1 + A 2 + A 3 + A 4 A1+A2+A3+A4 14 + 18 + 22 + 26 + . . . + 86 What is the value of sum?

postnew [5]

Answer:

950

Step-by-step explanation:

The common difference is 4, so the general term can be written:

... an = 14 + 4(n -1)

The value of n for the last term is ...

... 86 = 14 + 4(n -1) . . . . . the computation for the last term, 86

... 72 = 4(n -1) . . . . . . . . . subtract 14

... 18 = n -1 . . . . . . . . . . . divide by 4

... 19 = n . . . . . . . . . . . . . add 1

Your series has 19 terms. The first term is 14 and the last is 86, so the average term is (14+86)/2 = 50. Since there are 19 terms, the sum of them is ...

... 19×50 = 950

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

4 years ago

Solve pls brainliest

VikaD [51]

Answer: -5.27 + (-3.32) = -8.59 and -5.6 + 27.4 = 21.8

Step-by-step explanation: Have a nice day bro!✌️

6 0

2 years ago

Describe the rule for the rotation.

finlep [7]

Answer:

Step-by-step explanation:

If you draw a line from the origin (0,0) to L ( the original point ) and a different line from the origin to the image L' you can see the angle of rotation as being

90 degrees and that the rotation is clockwise.

the rule is (x, y) become ( y, -x)

4 0

3 years ago