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Ghella [55]
4 years ago
12

Henry divided his socks into five equal groups. Let s represent the total number of socks. Which expression and solution represe

nt the number of socks in each group if s = 20?
Mathematics
1 answer:
WINSTONCH [101]4 years ago
5 0

Answer:

20/5=number in each group. It can also be S/5

Step-by-step explanation:

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What are the coordinates
anzhelika [568]

Answer:

(1.5,1)

Step-by-step explanation:

I don't know how to explain. I used GeoGebra and I put the coordinate down and then made lines. Line AC and line BD. Then I put a point on where it intersects and it comes out to (1.5,1). You can also use distance formula, but that is more work.

5 0
3 years ago
Find a pattern for each sequence 0,1, 0.01, 0.001,.....
Ghella [55]

Answer:

Divided by 10 every time

Step-by-step explanation:

/10 is the pattern

8 0
2 years ago
Sally is completely unprepared for a three-question multiple-choice pop quiz, so she randomly guesses the answer to each questio
timurjin [86]
There are 3 questions and 4 choices.
So she has 1/4 of a chance of getting each question right.
Now if you multiply all her odds 1/4 x 1/4 it would become 1/16. Therefore she has 1/16 of a chance of getting a 2/3
8 0
4 years ago
How do I figure out the volume of this image
-BARSIC- [3]
The volume of a rectangular prism can be found from the formula V=B•h
The base is the triangle
The area of the triangle can found using A=b•1/2h
13•1/2(8)=52
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468in^2
8 0
3 years ago
Find a particular solution to the nonhomogeneous differential equation y'' + 4 y = cos(2x) + sin(2x).
alukav5142 [94]
The characteristic solution follows from solving the characteristic equation,

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

so that

y_c=C_1\cos2x+C_2\sin2x

A guess for the particular solution may be a\cos2x+b\sin2x, but this is already contained within the characteristic solution. We require a set of linearly independent solutions, so we can look to

y_p=ax\cos2x+bx\sin2x

which has second derivative

{y_p}''=(-4ax+4b)\cos2x+(-4bx-4a)\sin2x

Substituting into the ODE, you have

y''+4y=\cos2x+\sin2x
\implies4b\cos2x-4a\sin2x=\cos2x+\sin2x
\implies\begin{cases}4b=1\\-4a=1\end{cases}\implies a=-\dfrac14,b=\dfrac14

Therefore the particular solution is

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

Note that you could have made a more precise guess of

y_p=(a_1x+a_0)\cos2x+(b_1x+b_0)\sin2x

but, of course, any solution of the form a_0\cos2x+b_0\sin2x is already accounted for within y_c.
6 0
3 years ago
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