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Novosadov [1.4K]
3 years ago
10

What is 1/2+1/2+1/2=

Mathematics
1 answer:
katrin2010 [14]3 years ago
8 0

Answer:

the answer would be 2 and 1/2

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Please I need this before 12pm
luda_lava [24]

Answer:

The required formula is:

                                                          {\displaystyle \ a_{n}=a_{1}+(n-1)d}

Step-by-step explanation:

The total number of squares of the the first term = 4

The total number of squares of the the second term = 7

The total number of squares of the the third term = 10

so,

a_1=4

a_2=7

a_3=10

Finding the common difference d

d=a_3-a_2=10-7=3

d=a_2-a_1=7-4=3

As the common difference 'd' is same, it means the sequence is in arithmetic.

So

If the initial term of an arithmetic progression is {\displaystyle a_{1}} and the common difference of successive members is d, then the nth term of the sequence (a_n) is given by:

                         {\displaystyle \ a_{n}=a_{1}+(n-1)d}

Therefore, the required formula is:

                                                          {\displaystyle \ a_{n}=a_{1}+(n-1)d}

3 0
3 years ago
Which statement best explains how Italian society was affected by the Renaissance?
Tresset [83]

italian society entered a new era of artistic and academic achievement

4 0
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Someone please help !! picture shown
Aleks [24]
Correct me if im wrong but the answer is (B) hope this helps!
5 0
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THIS IS PROBABLY MY LAST QUESTION!!!!!!!!
Lady bird [3.3K]

Answer:

2.c 3.a 4.b

Step-by-step explanation:

just looked at it carefully

6 0
3 years ago
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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
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