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IRINA_888 [86]
3 years ago
14

Graph the function y=-x+3 using inputs of -1,0,1, and 2

Mathematics
1 answer:
galina1969 [7]3 years ago
8 0

Answer:

Given function:

y=-x+3

To graph the line using inputs -1,0,1 and 2.

Solution:

In order to find the points to be plot on graph we will plugin the input values as x in the given function.

1) When x=-1

y=-(-1)+3

y=1+3

y=4

Point (-1,4)

2) When x=0

y=-(0)+3

y=0+3

y=3

Point (0,3)

3) When x=1

y=-(1)+3

y=-1+3

y=2

Point (1,2)

4) When x=2

y=-(2)+3

y=-2+3

y=1

Point (2,1)

The points to be plot in the graph are:

(-1,4),(0,3),(1,2),(2,1)

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Find the solution of the differential equation that satisfies the given initial condition. y' tan x = 3a + y, y(π/3) = 3a, 0 &lt
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Answer:

y(x)=4a\sqrt{3}* sin(x)-3a

Step-by-step explanation:

We have a separable equation, first let's rewrite the equation as:

\frac{dy(x)}{dx} =\frac{3a+y}{tan(x)}

But:

\frac{1}{tan(x)} =cot(x)

So:

\frac{dy(x)}{dx} =cot(x)*(3a+y)

Multiplying both sides by dx and dividing both sides by 3a+y:

\frac{dy}{3a+y} =cot(x)dx

Integrating both sides:

\int\ \frac{dy}{3a+y} =\int\cot(x) \, dx

Evaluating the integrals:

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Where C1 is an arbitrary constant.

Solving for y:

y(x)=-3a+e^{C_1} sin(x)

e^{C_1} =constant

So:

y(x)=C_1*sin(x)-3a

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Solving for C1:

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Therefore:

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