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inessss [21]
3 years ago
7

Which of the following is an equation of the line in the graph?

Mathematics
1 answer:
zlopas [31]3 years ago
3 0
<h3>Answer:</h3>

B. 3x +y = 4

<h3>Step-by-step explanation:</h3>

It is perhaps easiest to simply try the equations to see which one works.

For x=0, there are two different kinds of answers:

... A and C: -y = 4

... B and D: y = 4

Since we know y=4 when x=0 (from the point (0, 4)), we can eliminate choices A and C.

___

Using the point (1, 1), you can try choices B and D to see which works:

... B: 3·1 +1 = 4 . . . . true (put 1 where x and y are in the equation)

... D: -3·1 +1 = -2 = 4 . . . . false

The appropriate choice is the equation of B: 3x +y = 4.

_____

<em>Derive the equation from the given points</em>

There are several ways you can derive the equation. Since you have the y-intercept (the point with x=0), you can use the slope-intercept form to start.

The slope (m) is ...

... m = (change in y)/(change in x) = (4 -1)/(0 -1)

... m = -3

We know the y-intercept (b) is 4, so the slope-intercept form of the equation is ...

... y = mx +b

... y = -3x +4

Adding 3x puts this in standard form:

... 3x +y = 4

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Complete factorization for -20x - 5y
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-5 (4x+y)

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7 0
3 years ago
Determine whether the equation is exact. If it​ is, then solve it. e Superscript t Baseline (7 y minus 3 t )dt plus (2 plus 7 e
umka21 [38]

Answer:

F(t,y)=(2+7e^t)y+3(1-t)e^t +C

Step-by-step explanation:

You have the following differential equation:

e^t(7y-3t)dt+(2+7e^t)dy=0

This equation can be written as:

Mdt+Ndy=0

where

M=e^t(7y-3t)\\\\N=(2+7e^t)

If the differential equation is exact, it is necessary the following:

\frac{\partial M}{\partial y}=\frac{\partial N}{\partial t}

Then, you evaluate the partial derivatives:

\frac{\partial M}{\partial y}=\frac{\partial}{\partial t}e^t(7y-3t)\\\\\frac{\partial M}{\partial t}=7e^t\\\\\frac{\partial N}{\partial t}=\frac{\partial}{\partial t}(2+7e^t)\\\\\frac{\partial N}{\partial t}=7e^t\\\\\frac{\partial M}{\partial t} = \frac{\partial N}{\partial t}

The partial derivatives are equal, then, the differential equation is exact.

In order to obtain the solution of the equation you first integrate M or N:

F(t,y)=\int N \partial y = (2 +7e^t)y+g(t)        (1)

Next, you derive the last equation respect to t:

\frac{\partial F(t,y)}{\partial t}=7ye^t+g'(t)

however, the last derivative must be equal to M. From there you can calculate g(t):

\frac{\partial F(t,y)}{\partial t}=M=(7y-3t)e^t=7ye^t+g'(t)\\\\g'(t)=-3te^t\\\\g(t)=-3\int te^tdt=-3[te^t-\int e^tdt]=-3[te^t-e^t]

Hence, by replacing g(t) in the expression (1) for F(t,y) you obtain:

F(t,y)=(2+7e^t)y+3(1-t)e^t +C

where C is the constant of integration

8 0
3 years ago
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