can the area of a square be expressed as a prime number if its side length is expressed as a natural number?

Mathematics

2 answers:

Lerok [7]3 years ago

7 0

not possible because the area of a square is always a square of it's length and a square cannot be a prime number

Airida [17]3 years ago

7 0

Answer:

Step-by-step explanation:

Because the area of a square is always a square of its length and a square can't be prime.

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Help pls TNX............

Nitella [24]

Answer:

Step-by-step explanation:

3 0

3 years ago

All are equivalent except x<-2, x-2<4, 2x<-4, x-2<-4 ?

m_a_m_a [10]

To figure out which is not equivalent to the others,, we must solve each option provided. x < -2 is already solved,, so there is no need to do any work that option.
The first step for solving x - 2 < 4 is to move the constant to the right side and then change its sign.
x < 4 + 2
Now add the numbers together to get your final answer.
x < 6
This means that we have one option that equals x < -2 and one option that equals x < 6.
Let's now solve 2x < -4 to see what that one equals. In order to solve this,, we need to divide both sides of the inequality by 2.
x < -2
Now we can see that it looks like all of the expressions are equivalent except for x - 2 < 4. Before we can confirm this though,, let's solve for x - 2 < -4. The first step for solving this is to move the constant to the right side and change its sign.
x < -4 + 2
Now calculate the sum of these two numbers to get your final answer.
x < -2
This tells us that all of the options are equivalent except for x - 2 < 4,, or option B.
Let me know if you have any further questions.
:)

4 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

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