Can you guyz help me with this question step by step please ☺️

Mathematics

1 answer:

zlopas [31]3 years ago

7 0

2 1/2 x 3 = 7 1/2 = 7 4/8

10 11/8
- 7 4/8
3 7/8
A is the correct answer

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Which of the following is the product of the rational expression shown below

wolverine [178]

Option C: $\frac{x^{2}-9}{x^{2}-4}$ is the product of the rational expression.

Explanation:

The given rational expression is $\frac{x+3}{x+2} \cdot \frac{x-3}{x-2}$

We need to determine the product of the rational expression.

<u>Product of the rational expression:</u>

Let us multiply the rational expression to determine the product of the rational expression.

Thus, we have;

$\frac{(x+3)(x-3)}{(x+2)(x-2)}$

Let us use the identity $(a+b)(a-b)=a^2-b^2$ in the above expression.

Thus, we get;

$\frac{x^{2} -3^2}{x^{2} -2^2}$

Simplifying the terms, we get;

$\frac{x^{2}-9}{x^{2}-4}$

Thus, the product of the rational expression is $\frac{x^{2}-9}{x^{2}-4}$

Hence, Option C is the correct answer.

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

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