All categories

3 years ago

What is the answer to 125÷3

Mathematics

2 answers:

Sunny_sXe [5.5K]3 years ago

5 0

Answer:

41.6666667?? I didnt know if you needed is simplified .-.

labwork [276]3 years ago

4 0

Answer:

41.6 repeating .

Step-by-step explanation:

125 divided by 3 is 41.6 with the 6 repeating .

You might be interested in

Describe which strategy you would use to divide 48 by 8

Whitepunk [10]

You can repeat 8 till you get to 48 and count how many 8's you needed to get there.

This is one of the many strategys but there are better ones.

3 0

3 years ago

Read 2 more answers

(PLZ help)

k0ka [10]

2.because
5.50x4=22
2.25x2=4.50
that will equal 26.50 but when you add the money she already had saved up which is 5.25
you will get 31.75

4 0

2 years ago

Read 2 more answers

900,000 divided by 3,000

mina [271]

300.................................

8 0

3 years ago

Read 2 more answers

Find the simplified form of each expression. (3/5y^4)^-2

Keith_Richards [23]

25/9y^8
I believe its this

3 0

2 years ago

Por favor ayuda con estw problema de la transformada de ls derivada<br> Y''-6y+9y=t y(0)=0 y' (0)=1

Hunter-Best [27]

Parece que refieres a la transformada de Laplace. Aplica la transformada a los lados ambos de la ecuación diferencial:

$\mathcal L_s\{y''(t)-6y'(t)+9y(t)\}=\mathcal L_s\{t\}$

Denota por $Y(s)=\mathcal L_s\{y(t)\}$ la transformada de $y(t)$ . En el lado izquierdo obtenemos

$\mathcal L_s\{y''(t)-6y'(t)+9y(t)\}=\mathcal L_s\{y''(t)\}-6\mathcal L_s\{y'(t)+9\mathcal L_s\{y(t)\}$

$\mathcal L_s\{y''(t)-6y'(t)+9y(t)\}=(s^2Y(s)-sy(0)-y'(0))-6(sY(s)-y(0))+9Y(s)$

$\mathcal L_s\{y''(t)-6y'(t)+9y(t)\}=(s^2-6s+9)Y(s)-1$

$\mathcal L_s\{y''(t)-6y'(t)+9y(t)\}=(s-3)^2Y(s)-1$

y a la derecha,

$\mathcal L_s\{t\}=\dfrac1{s^2}$

Ahora resuelve para $Y(s)$ :

$(s-3)^2Y(s)-1=\dfrac1{s^2}$

$(s-3)^2Y(s)=1+\dfrac1{s^2}$

$Y(s)=\dfrac{s^2+1}{s^2(s-3)^2}$

Expande la fracción por la derecha en fracciones parciales; busque $a,b,c,d$ tal que

$\dfrac{s^2+1}{s^2(s-3)^2}=\dfrac as+\dfrac b{s^2}+\dfrac c{s-3}+\dfrac d{(s-3)^2}$

$s^2+1=as(s-3)^2+b(s-3)^2+cs^2(s-3)+ds^2$

$s^2+1=(a+c)s^3+(-6a+b-3c+d)s^2+(9a-6b)s+9b$

Emparejando los términos con igual grado da el sistema con soluciones

$\begin{cases}a+c=0\\-6a+b-3c+d=1\\9a-6b=0\\9b=1\end{cases}\implies a=\dfrac2{27},b=\dfrac19,c=-\dfrac2{27},d=\dfrac{10}9$

Pues

$Y(s)=\dfrac2{27}\dfrac1s+\dfrac19\dfrac1{s^2}-\dfrac2{27}\dfrac1{s-3}+\dfrac{10}9\dfrac1{(s-3)^2}$

y tomar la transformada inversa es trivial. Obtenemos

$\mathcal L^{-1}_t\{Y(s)\}=\dfrac2{27}\mathcal L^{-1}_t\left\{\dfrac1s\right\}+\dfrac19\mathcal L^{-1}_t\left\{\dfrac1{s^2}\right\}-\dfrac2{27}\mathcal L^{-1}_t\left\{\dfrac1{s-3}\right\}+\dfrac{10}9\mathcal L^{-1}_t\left\{\dfrac1{(s-3)^2}\right\}$

$\boxed{y(t)=\dfrac2{27}+\dfrac t9-\dfrac2{27}e^{3t}+\dfrac{10}9te^{3t}}$

8 0

3 years ago

What is the answer to 125÷3​

What is the answer to 125÷3