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Eddi Din [679]
4 years ago
7

Please help me with this question!

Mathematics
1 answer:
nikitadnepr [17]4 years ago
4 0

Answer:

1. 8/9x11/7

2. 8/9x10/3

3. 7x15/1

4.7x10/3

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As x approaches positive infinity, which of the following describes the end behavior for
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If it is 1:10 in virginia right now and its 6:10 in england how many hours are they apart
dalvyx [7]

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5 hours

Step-by-step explanation:

6:10 - 1:10 = 5

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Serggg [28]
9a+5b you need to distribute the numbers in parenthesis
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Read 2 more answers
Please help me it would mean the world. Extra points!
tensa zangetsu [6.8K]
A = ((B+b)*h)/2
A = ((6+4)*3)/2
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7 0
3 years ago
: Show that the solution of the differential equation: = − − − − − is of the form: + + ( − ) = + , When = and =
Serhud [2]

Answer:

y = \tan(x + \frac{x^2}{2})

Step-by-step explanation:

Poorly formatted question; The complete question requires that we prove that y=\tan(x+\frac{x\²}{2})

When

\frac{dy}{dx} =1+xy\²+x+y\² and y(0)=0  

We have:

\frac{dy}{dx} =1+xy\²+x+y\²

Rewrite as:

\frac{dy}{dx} =1+x+xy\²+y\²

Factorize

\frac{dy}{dx} = (1+x)+y\²(x+1)

Rewrite as:

\frac{dy}{dx} = (1+x)+y\²(1+x)

Factor out 1 + x

\frac{dy}{dx} = (1+y\²)(1+x)

Multiply both sides by \frac{dx}{1 + y^2}

\frac{dy}{1+y\²} = (1+x)dx

Integrate both sides

\int \frac{dy}{1+y\²} = \int (1+x)dx

Rewrite as:

\int \frac{1}{1+y\²} dy = \int (1+x)dx

Integrate the left-hand side

\int \frac{1}{1+y\²} dy = \tan^{-1}y

Integrate the right-hand side

\tan^{-1}y = x + \frac{x^2}{2} + c

y(0)=0 implies that: (x,y) = (0,0)

So:

\tan^{-1}y = x + \frac{x^2}{2} + c becomes

\tan^{-1}(0) = 0 + \frac{0^2}{2} + c

This gives:

0 = 0 +0 + c

0 =c

c = 0

The equation \tan^{-1}y = x + \frac{x^2}{2} + c becomes

\tan^{-1}y = x + \frac{x^2}{2} + 0

\tan^{-1}y = x + \frac{x^2}{2}

Take tan of both sides

y = \tan(x + \frac{x^2}{2}) --- Proved

8 0
3 years ago
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