All categories

3 years ago

Math help plz

Mathematics

1 answer:

bija089 [108]3 years ago

8 0

Answer:

I guess this is the answer hope it helps

You might be interested in

Can someone answer this for me please

Nana76 [90]

Answer:

Step-by-step explanation:

6 0

3 years ago

A wise women once said, “400 reduced by twice my age is 264.” What is her age?

KonstantinChe [14]

Answer:

My age is 78 years.

Explanation:

Define my age first as a variable.

Let my age be

years.

Reduced by means subtracted.

Subtract twice my age from 400 to get 244:

400

−

244

400

−

244

←

make x positive and isolate it.

156

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

If the first and the last terms of an arithmetic series are 10 and 62, show that the sum of the series varies directly as the nu

Nitella [24]

Answer:

10+^@=72_^) + the series

Step-by-step explanation:

6 0

3 years ago

A swimming class has 20 students. Only 8 students know how to dive off

Allisa [31]

Answer: 40%

Step-by-step explanation:

To write 8/20 as a percent have to remember that 1 equal 100% and that what you need to do is just to multiply the number by 100 and add at the end symbol % . 8/20 * 100 = 0.4 * 100 = 40% And finally we have: 8/20 as a percent equals 40%

6 0

3 years ago

3y''-6y'+6y=e*x sexcx

Simora [160]

From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

$3r^2-6r+6=0\iff r^2-2r+2=0$

which has roots at $r=1\pm i$ . This admits the two fundamental solutions

$y_1=e^x\cos x$
$y_2=e^x\sin x$

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

$y_p=u_1y_1+u_2y_2$

where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

$u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx$
$u_1=\dfrac13\ln|\cos x|$

$u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx$
$u_2=\dfrac13x$

Therefore the particular solution is

$y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

7 0

3 years ago