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2 years ago

How to write 1 whole and 2 fiths as a fraction only

Mathematics

2 answers:

BaLLatris [955]2 years ago

7 0

$1 \frac{2}{5} = \frac{(1\cdot5) + 2}{5} = \frac{5+2}{5} = \boxed{\bf{\frac{7}{5}}}$

Juliette [100K]2 years ago

6 0

1 2/5
(5x1)+2=7 soo as an improper fraction that would be 7/5

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Need help please. I don't know

Natali [406]

$f(3)=3+1=4$

6 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

Help. Explain good for brainliest

Nimfa-mama [501]

The measure of m∠J=90°, m∠G=55° and m∠H=35°

Step-by-step explanation:

Inscribed angles subtended by a diameter are right angles. Hence angle GJH =90°

The diameter GH and chord GJ intercepts an arc with a measure of 70°.The measure of an arc of a circle is equal to the measure of the central angle that intercepts the arc. Hence angle GOJ=70°. Radius of circle GO=OJ, which means triangle GOJ is an isosceles triangle thus ∠OJG=∠OGJ =(180° -70°) /2 = 55°. m∠G=55°

m∠H = (180°-(90°+55°) ) = 180° - 145° =35°

Learn More

Angles subtended by a diameter :brainly.com/question/10345730

Keywords : measures, angles, diameter, chord, intercepts, arc

#LearnwithBrainly

8 0

3 years ago

Can u guys plz help me with my last 2 questionsssss. It’s 26 and 27

skad [1K]

I think 27 is 1350 since you just multiply and add

4 0

2 years ago

Given f(x)=2x-4 and g(x)=-3x^2+5x determine f(3)-g(2)

victus00 [196]

Answer:

Step-by-step explanation:

Evaluate f(3) and g(2)

f(3) = 2(3) - 4 = 6 - 4 = 2

g(2) = - 3(2)² + 5(2) = - 3(4) + 10 = - 12 + 10 = - 2

Then

f(3) - g(2) = 2 - (- 2) = 2 + 2 = 4

8 0

3 years ago