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Sergio039 [100]
3 years ago
10

Solve the initial value problem. dy/dt = 1 + 6/t , t > 0, y = 8 when t = 1

Mathematics
1 answer:
nordsb [41]3 years ago
5 0
\displaystyle\frac{dy}{dt} = 1 + \frac{6}{t}\ \Rightarrow\ dy = \left( 1 + \frac{6}{t}\right) dt\ \Rightarrow\int 1\, dy = \int \left( 1 + \frac{6}{t}\right) dt \ \Rightarrow \\ \\
y = t + 6\ln|t| + C. \text{ But $t\ \textgreater \ 0$ so }y = t + 6\ln t + C. \\ \\ 
y(1) = 8\ \Rightarrow\ 8 = 1 + 6 \ln 1 + C \ \Rightarrow\ C = 7 \text{ so } \\ \\
y(t) = t + 6\ln t + 7
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a line passes through the point (-9, 1) and had a slope of -2/3. Write an equation in point-slope form for this line
coldgirl [10]

Answer:

y-1=-2/3(x+9)

Step-by-step explanation:

y-1=-2/3(x-(-9))

y-1=-2/3(x+9)

7 0
3 years ago
Read 2 more answers
The numerator of a given fraction is 4 less than the denominator. if 3 is subtracted from the numerator and 5 is added to the de
Sergeu [11.5K]

Answer:

Step-by-step explanation:Given that the numerator of a given fraction is 4 less than its denominator.

Also given that 3 is subtracted from the numerator and 5 is added to the denominator, the fraction becomes one by fourth .

Let the fraction be  

Since the numerator of a given fraction is 4 less than its denominator we have,

Numerator=Denominator-4

⇒ a=b-4

Since 3 is subtracted from the numerator and 5 is added to the denominator, the fraction becomes one by fourth we have

4(a-3)=1(b+5)

4a-12=b+5

4a-b=17

4(b-4)-b=17   ( ∵ a=b-4)

4b-16-b=17

3b=17+16

3b=33

⇒ b=11

Now put b=11 in a=b-4 we get

a=11-4

⇒  the fraction is a/b=7/11

3 0
3 years ago
What is the y- intercept of: y=-2x
shusha [124]

Answer:

it is Zero 0

Step-by-step explanation:

7 0
3 years ago
Can someone help w/ 3 and 4 please??
Ymorist [56]
Yep that looks about right. I would say that works. Good job. You're doing great
3 0
3 years ago
Can someone help me with this​
SashulF [63]

Answer:

\angle NRQ =85 \degree

Given:

\angle RPQ = 45 \degree \\  \angle PQR = (6x + 4)\degree \\  \angle NRQ = (15x - 5)\degree

Step-by-step explanation:

Property used: <em>An exterior angle of a triangle is equal to the sum of the opposite int</em><em>e</em><em>rior angles.</em>

<em>=  >  \angle RPQ +  \angle PQR =  \angle NRQ \\  \\  =  > 45 \degree + (6x + 4)\degree = (15x - 5)\degree \\  \\  =  > 45\degree + 6x\degree + 4\degree = 15x\degree - 5\degree \\  \\  =  > 49\degree  + 5 \degree= 15x\degree - 6x\degree \\  \\  =  > 54\degree = 9x\degree \\  \\  =  > 9x\degree = 54\degree \\  \\  =  > x\degree =  (\frac{54}{9} )\degree \\  \\  =  > x\degree = 6\degree \\  \\ Putting \: value \: of \: x \: in \:  \angle NRQ  \\   \\  =  > \angle NRQ = (15x - 5)\degree \\  \\  =  >  \angle NRQ = (15 \times 6 - 5) \degree \\  \\  =  >  \angle NRQ =85 \degree</em>

4 0
3 years ago
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