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Jet001 [13]
3 years ago
10

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:

No

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:

6

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
7 0
3 years ago
If x = -6 and y = -2, evaluate the expression.<br> 3x - 10<br> y-5
Bond [772]
Is it
3x-10= 3(-6)-10= -28
Y-5= -2-5=-7
3 0
3 years ago
1. what are the domain and range of the function?
natta225 [31]

1) domain of the first is ( R ) because of its root(3) and its range is (R) because of its root again

2)and domain of the last , is [0,+infinity)

and for its range , you can draw it such as picture and will be [0,- infinity)

7 0
3 years ago
Read 2 more answers
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