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Katena32 [7]
3 years ago
8

Tom is thinking of two numbers,a and b,where a is a positive and b is negative. write to inequalities that Tim can use to compar

e a and b
Explain
Mathematics
1 answer:
Advocard [28]3 years ago
8 0

Answer:a﹥0﹥b

Step-by-step explanation:

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What is the value of X
ExtremeBDS [4]
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\dfrac{2}{3}=\dfrac{x+3}{12}\\
3(x+3)=24\\
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3 years ago
2 points) Sometimes a change of variable can be used to convert a differential equation y′=f(t,y) into a separable equation. One
Stells [14]

y'=(t+y)^2-1

Substitute u=t+y, so that u'=y', and

u'=u^2-1

which is separable as

\dfrac{u'}{u^2-1}=1

Integrate both sides with respect to t. For the integral on the left, first split into partial fractions:

\dfrac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)=1

\displaystyle\int\frac{u'}2\left(\frac1{u-1}-\frac1{u+1}\right)\,\mathrm dt=\int\mathrm dt

\dfrac12(\ln|u-1|-\ln|u+1|)=t+C

Solve for u:

\dfrac12\ln\left|\dfrac{u-1}{u+1}\right|=t+C

\ln\left|1-\dfrac2{u+1}\right|=2t+C

1-\dfrac2{u+1}=e^{2t+C}=Ce^{2t}

\dfrac2{u+1}=1-Ce^{2t}

\dfrac{u+1}2=\dfrac1{1-Ce^{2t}}

u=\dfrac2{1-Ce^{2t}}-1

Replace u and solve for y:

t+y=\dfrac2{1-Ce^{2t}}-1

y=\dfrac2{1-Ce^{2t}}-1-t

Now use the given initial condition to solve for C:

y(3)=4\implies4=\dfrac2{1-Ce^6}-1-3\implies C=\dfrac3{4e^6}

so that the particular solution is

y=\dfrac2{1-\frac34e^{2t-6}}-1-t=\boxed{\dfrac8{4-3e^{2t-6}}-1-t}

3 0
3 years ago
What is the value of t in the equation 3(2t + 5) = 5t + 25?
11Alexandr11 [23.1K]
<span>3(2t + 5) = 5t + 25
6t + 15 = 5t + 25
6t - 5t = 25 - 15
t = 10</span>
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3 years ago
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You can try by plugging each ordered pair and seeing if the equation comes out true.

(5, 1)

2 * 5 - 1 = 9

That's correct so C is the correct answer.

6 0
3 years ago
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