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leva [86]
3 years ago
8

Find the 3rd term in the expansion of (a + b)^7 in simplest form.

Mathematics
1 answer:
sergey [27]3 years ago
5 0

9514 1404 393Answer:

  21·a^5·b^2

Step-by-step explanation:

The k-th term is ...

  C(7, k)·a^(7-k)·b^k . . . . . where k counts from 0 to n

Then k=2 for the third term. The coefficient of it is ...

  C(7, 2) = 7!/(2!(7-2)!) = 7·6/(2·1) = 21

The third term is ...

  21·a^5·b^2

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A loaf of bread weighs about 1pound. Chose all of the metric measures that are less than 1pound. 1kg 0.5kg 500g 450g 100g
wel

Answer:  E) 450g and F) 100g

Step-by-step explanation:

A pound is approximately <em>0.4535 kilogrammes</em>, so <em>453,5 grammes</em>. With this information, we can determine that answers E (450 grammes) and F ( 100 grammes) are the only correct ones.

3 0
3 years ago
The equation y=ax describes the graph of a line.if the value of a is positive,the line
ahrayia [7]
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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
2x+4≥24 how do I solve this problem
Katen [24]
First subtract 4 then divide 2 so u can isolate x
3 0
4 years ago
-56 ÷ c = 7, c =<br> pls help me
Rus_ich [418]

Answer:

-392

Step-by-step explanation:

-56 x 7=-392

-392 / -56=7

7 0
3 years ago
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