All categories

3 years ago

I need help with it pls help I will really appreciate it !

Mathematics

1 answer:

GuDViN [60]3 years ago

8 0

Answer:

try with this link https://es.symbolab.com/solver/square-roots-calculator

Step-by-step explanation:

You might be interested in

2k•squared• + 7 - 15<br> factoring ax•squared•+bx+c

Rainbow [258]

Calculate Δ = b^2 -4ac => Δ = 169 => k1 = 3/2; k2 = -5 => 2k^2 + 7k - 15 = 2*(k -3/2)(k+5) = (2k - 3)(k+5);

6 0

4 years ago

Solve the initial value problem 2ty" + 10ty' + 8y = 0, for t > 0, y(1) = 1, y'(1) = 0.

Eva8 [605]

I think you meant to write

$2t^2y''+10ty'+8y=0$

which is an ODE of Cauchy-Euler type. Let $y=t^m$ . Then

$y'=mt^{m-1}$

$y''=m(m-1)t^{m-2}$

Substituting $y$ and its derivatives into the ODE gives

$2m(m-1)t^m+10mt^m+8t^m=0$

Divide through by $t^m$ , which we can do because $t\neq0$ :

$2m(m-1)+10m+8=2m^2+8m+8=2(m+2)^2=0\implies m=-2$

Since this root has multiplicity 2, we get the characteristic solution

$y_c=C_1t^{-2}+C_2t^{-2}\ln t$

If you're not sure where the logarithm comes from, scroll to the bottom for a bit more in-depth explanation.

With the given initial values, we find

$y(1)=1\implies1=C_1$

$y'(1)=0\implies0=-2C_1+C_2\implies C_2=2$

so that the particular solution is

$\boxed{y(t)=t^{-2}+2t^{-2}\ln t}$

# # #

Under the hood, we're actually substituting $t=e^u$ , so that $u=\ln t$ . When we do this, we need to account for the derivative of $y$ wrt the new variable $u$ . By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm dy}{\mathrm du}$

Since $\frac{\mathrm dy}{\mathrm dt}$ is a function of $t$ , we can treat $\frac{\mathrm dy}{\mathrm du}$ in the same way, so denote this by $f(t)$ . By the quotient rule,

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac ft\right]=\dfrac{t\frac{\mathrm df}{\mathrm dt}-f}{t^2}$

and by the chain rule,

$\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\mathrm df}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm df}{\mathrm du}$

where

$\dfrac{\mathrm df}{\mathrm du}=\dfrac{\mathrm d}{\mathrm du}\left[\dfrac{\mathrm dy}{\mathrm du}\right]=\dfrac{\mathrm d^2y}{\mathrm du^2}$

so that

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm dy}{\mathrm du}}{t^2}=\dfrac1{t^2}\left(\dfrac{\mathrm d^2y}{\mathrm du^2}-\dfrac{\mathrm dy}{\mathrm du}\right)$

Plug all this into the original ODE to get a new one that is linear in $u$ with constant coefficients:

$2t^2\left(\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm d y}{\mathrm du}}{t^2}\right)+10t\left(\dfrac{\frac{\mathrm dy}{\mathrm du}}t\right)+8y=0$

$2y''+8y'+8y=0$

which has characteristic equation

$2r^2+8r+8=2(r+2)^2=0$

and admits the characteristic solution

$y_c(u)=C_1e^{-2u}+C_2ue^{-2u}$

Finally replace $u=\ln t$ to get the solution we found earlier,

$y_c(t)=C_1t^{-2}+C_2t^{-2}\ln t$

4 0

4 years ago

PLEASE HELP whats the answer for this problem

MArishka [77]

Answer:

it is 20 me and that other kid did the same thing

Step-by-step explanation:

5x+12 + 3x+8 =180

8x=180-20

8x=160

hope that helps brainly plz

x=20

4 0

3 years ago

Mrs. Myles wants to know if her students prefer multiple choice or short answer tests. What question would be MOST appropriate t

djverab [1.8K]

Answer: The answer is multiple choice questions.

Step-by-step explanation: Given that Mrs. Myles wants to know if her students prefer multiple choice or short answer tests.

According to me, Mrs. Myles should ask her students multiple choice questions, because in these type of questions, we get information not only about the correct option but it also reminds us what the incorrect options are related to.

So, these questions are better than short answers which are related to only one topic rather than three or four in the queue for multiple choice questions.

Thus, Mrs. Myles should ask her students multiple choice questions.

5 0

4 years ago

Read 2 more answers

The correlation analysis assumes that the measurements have a bivariate normal distribution in the population. Select all of the

hoa [83]

Answer:

Bruh i got correlation hw that i posted on here too and nobody helped lol.

Step-by-step explanation:

8 0

3 years ago