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Ket [755]
3 years ago
9

What is the perimeter of the shape shown in the graph below?

Mathematics
1 answer:
IrinaVladis [17]3 years ago
7 0
The answer is D)15.45 because if u count the boxes it’s around 10 then about a couple more with the added ones that are in half so the best answer that would make sense is that. Hope it helps!!
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Does any one know the answer please​
Fittoniya [83]

Answer:

Slope intercept form

Step-by-step explanation:

y = Mx + B

M = slope - 7/11

B = y intercept - 5

5 0
3 years ago
Hii guys I am riyaz aly​
lesya [120]
Hello im Esmerallda
How are u
6 0
3 years ago
6x - 170 = -4x + 110​
strojnjashka [21]

Answer:

=-110

Step-by-step explanation:

:)

7 0
3 years ago
Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).
I am Lyosha [343]
Take the homogeneous part and find the roots to the characteristic equation:

y''+4y=0\implies r^2+4=0\implies r=\pm2i

This means the characteristic solution is y_c=C_1\cos2x+C_2\sin2x.

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form y_p=ax\cos2x+bx\sin2x. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With y_1=\cos2x and y_2=\sin2x, you're looking for a particular solution of the form y_p=u_1y_1+u_2y_2. The functions u_i satisfy

u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx

where W(y_1,y_2) is the Wronskian determinant of the two characteristic solutions.

W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2

So you have

u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx
u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x

u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx
u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x

So you end up with a solution

u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x

but since \cos2x is already accounted for in the characteristic solution, the particular solution is then

y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

so that the general solution is

y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x
7 0
3 years ago
Can someone tell me by number which ones are wrong please
trapecia [35]

Answer:idk

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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