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Serhud [2]
3 years ago
10

(4xy^2+6y)dx+(5yx^2+8x)dy=0 solve by using integrating factor

Mathematics
1 answer:
RSB [31]3 years ago
5 0
Are u sure about the second arrow?BTW
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Help!!!!!!<br> Please and thank you
fenix001 [56]

Answer:

1. A

2. B

3. A

Step-by-step explanation:

Hope this helps!

5 0
2 years ago
Find the product. (-3ab 2)3 27ab -23a2b2 -9a3b5 -27a3b6
horsena [70]

Answer:

Step-by-step explanation:

(-3 ab²)³=(-3)³(a)³(b²)³=-27 a³b^(2×3)=-27 a³b^6

8 0
3 years ago
Read 2 more answers
| x | &lt; 2 <br><br> solve the inequality and then graph the solution
anzhelika [568]

Answer:

Any number smaller than 2

Step-by-step explanation:

You would graph this by wrighting a solid, filled in circle with an arrow point toward the negative numbers, because those numbers are smaller than 2

3 0
2 years ago
A can of soda in 1972 costs $0.15, by 2012 it cost $1.35 find the rate at which the cost of a can of soda increases over this ti
fenix001 [56]

Answer:

800%

Step-by-step explanation:

<h2><u>Percentage increase</u></h2><h3><u>formula:</u></h3>

change/ original * 100

change is 1.35 - 0.15 = 1.2

original is the amount in the beginning which is 0.15

= 1.2/0.15 * 100

=800 %

<em><u>the percentage increase is by 800%</u></em>

8 0
2 years ago
If
Leno4ka [110]

Answer:

\frac{s^2-25}{(s^2+25)^2}

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given:\mathcal{L}[t \cos 5t]=(-1)F'(s) with F(s)=\mathcal{L}[\cos 5t].

Now, F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt. Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt.

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s).

Solving for F(s) on the last equation, F(s)=\frac{s}{s^2+25}, then the Laplace transform we were searching is -F'(s)=\frac{s^2-25}{(s^2+25)^2}

3 0
3 years ago
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