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topjm [15]
2 years ago
11

Find the common difference of the arithmetic sequence. 0, 0.4, 0.8, 1.2, . . .

Mathematics
1 answer:
irakobra [83]2 years ago
6 0
Each time it increases by 0.4

0 - 0.4 - 0.8 - 1.2 - 1.6 - 2.0 - 2.4 - 2.8 - 3.2 etc.
You might be interested in
Find two consecutive whole numbers that 7 lies between.
Wittaler [7]
The answer is actually 3 and 4.

\sqrt{7} lies between the two consecutive numbers 3 and 4.

I hope this helped! :))
5 0
3 years ago
Help please which expression could NOT be used to find the quotiend of 1,377 and 9
topjm [15]

Answer:

C

Step-by-step explanation:


7 0
2 years ago
D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x
MA_775_DIABLO [31]
The solution depends on the value of k. To make things simple, assume k>0. The homogeneous part of the equation is

\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0

and has characteristic equation

r^2-16k=0\implies r=\pm4\sqrt k

which admits the characteristic solution y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}.

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be y_p=ae^{4x}+be^x. Then

\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x

So you have

16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x
(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x

This means

16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}
b(1-16k)=30\implies b=\dfrac{30}{1-16k}

and so the general solution would be

y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x
8 0
2 years ago
I need help on this one guys :(
Sophie [7]

Answer:

the last one

Step-by-step explanation:

6 0
2 years ago
Item 15
Shtirlitz [24]

Answer:

28

Step-by-step explanation:

4*7= 28

when there is a variable (letter next to the number) next to the number like 4a it means 4 multiplied by a. since a is 7, 4 multiplied by 7 is 28.

8 0
2 years ago
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