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Korvikt [17]
3 years ago
9

Find the y-intercept of the following equation. Simplify your answer.

Mathematics
1 answer:
o-na [289]3 years ago
7 0

Answer:

<h2><em><u>Brainliest!</u></em></h2>

Step-by-step explanation:

the equation is y=mx+b

the b part is the y-intercept

so in your equation... the y=intercept would be 7/6

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4 0
3 years ago
Find the general solution of (x+3)y’=2y
gregori [183]

Answer:

y=C(x+3)^2

Step-by-step explanation:

We are given:

\displaystyle (x+3)y^\prime=2y

Separation of Variables:

\displaystyle \frac{1}{y}\frac{dy}{dx}=\frac{2}{x+3}

So:

\displaystyle \frac{dy}{y}=\frac{2}{x+3} \, dx

Integrate:

\displaystyle \int\frac{dy}{y}=\int\frac{2}{x+3}\, dx

Integrate:

\displaystyle \ln|y|=2\ln|x+3|+C

Raise both sides to e:

|y|=e^{2\ln|x+3|+C}

Simplify:

|y|=(e^{\ln|x+3|})^2\cdot e^C

So:

|y|=C|x+3|^2

Simplify:

y=\pm C(x+3)^2=C(x+3)^2

5 0
3 years ago
Help ASAP I need to get this question correct
strojnjashka [21]

Answer: The answer is NO

Step-by-step explanation:

8 0
3 years ago
Integrate t sec^2 (2t) dt
Neporo4naja [7]
F = t ⇨ df = dt 
dg = sec² 2t dt ⇨ g = (1/2) tan 2t 
⇔ 
integral of t sec² 2t dt = (1/2) t tan 2t - (1/2) integral of tan 2t dt 
u = 2t ⇨ du = 2 dt 

As integral of tan u = - ln (cos (u)), you get : 

integral of t sec² 2t dt = (1/4) ln (cos (u)) + (1/2) t tan 2t + constant 
integral of t sec² 2t dt = (1/2) t tan 2t + (1/4) ln (cos (2t)) + constant 
integral of t sec² 2t dt = (1/4) (2t tan 2t + ln (cos (2t))) + constant ⇦ answer 
8 0
4 years ago
Read 2 more answers
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