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3 years ago

Slope of line a is ½. What is the equation of a line that is parallel to line a? Y+½x=2 -2x+y+2=0 2x-2=y -½x+y-2=0

Mathematics

1 answer:

sasho [114]3 years ago

6 0

Answer:

The equation of a line that is parallel to line a is y=½x + 2, or what is the same -½x+y-2=0

Step-by-step explanation:

A line is a polynomial function of the first degree that has the following form:

y = m * x + b

where m is the slope of the line and b is the intercept with the Y axis. The slope is the inclination of the line with respect to the abscissa axis.

In this case you know that the slope of line a is ½.

Parallel lines are two or more lines in a plane that never intersect. Two lines are parallel if they have the same slope and different y-intercepts.

You have:

y+½x=2 → y= -½x +2
-2x+y+2=0 → y=2x -2
2x-2=y

-½x+y-2=0 → y=½x + 2

The line that has the same slope as line a is the last one, so <u><em>the equation of a line that is parallel to line a is y=½x + 2, or what is the same -½x+y-2=0</em></u>

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Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

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3 years ago

Can anyone explain this to me??

finlep [7]

It is the first option. DF = DE. It is referring to the lengths of each line, DE and DF look like they are the same. It is not a scalene triangle it is a right triangle. EF is way too long to be equal to DE or DF

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ololo11 [35]

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