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MrRissso [65]
2 years ago
14

There is a pair of parallel sides in the following shape. What is the area of the shape?

Mathematics
1 answer:
lesya [120]2 years ago
8 0

Answer:

63

Step-by-step explanation:

3x3=9. Ok? 9x7=63. That is how.             :)

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Solve for x: 3x^2+6x=2
vekshin1

Answer:

x = -1±sqrt(5/3)

Step-by-step explanation:

3x^2+6x=2

We will use completing the square

Factor out a 3 on the left

3(x^2 +2x) = 2

Divide each side by 3

3/3(x^2 +2x) = 2/3

(x^2 +2x) = 2/3

2/2 = 1 1^2 =1

Add 1 to each side

(x^2 +2x+1) = 2/3+1

(x+1) ^2 = 2/3 +3/3

(x+1)^2 = 5/3

Take the square root of each side

sqrt((x+1)^2) = ±sqrt(5/3)

x+1 = ±sqrt(5/3)

Subtract 1 from each side

x+1-1 = -1±sqrt(5/3)

x = -1±sqrt(5/3)


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The answer is 3,000.4%
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Add the following polynomials -4x^2+3xy-y^2 with -7x^2-Xy+6y^2 and x^2+4xy-2y^2
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Step-by-step explanation:


3 0
3 years ago
A container initially containing10 L of water in which there is 20 g of salt dissolved. A solution containing 4 g/L of salt is p
Kay [80]

Answer:

A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196

Step-by-step explanation:

For this case the solution flows at a rate of 2L/min and leaves at 1L/min. So then we can conclude the volume is given by V= 10 +t

Since the initial volume is 10 L and the volume increase at a rate of 1L/min.

For this case we can define A as the concentration for the salt in the container. And for this case we can set up the following differential equation:

\frac{dA}{dt}= 4 \frac{gr}{L} *2 \frac{L}{min} - \frac{A}{10+t}

Because at the begin we have a concentration of 8 gr/L and would be decreasing at a rate of \frac{A}{10+t}

So then we can reorder the differential equation like this:

\frac{dA}{dt} +\frac{A}{10+t} =8

We find the solution using the integration factor:

\mu = -\int \frac{1}{10+t} dt = -ln(10+t)

And then the solution would be given by:

A = e^{-ln (10+t)} (\int e^{\int \frac{1}{10+t} dt})

And if we simplify this we got:

A= \frac{1}{10+t} (c + \int (10 +t) 8 dt)

And after do the integral we got:

A= \frac{c}{10+t} +4 (10 +t)

And using the initial condition t=0 A= 20 we have this:

20 = \frac{c}{10} +40

c= -200

So then we have this function for the solution of A:

A= \frac{-200}{10+t} +4 (10 +t)

And now replacinf t= 40 we got:

A(40)= \frac{-200}{10+40} +4 (10 +40)=-4+200 = 196

8 0
3 years ago
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