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Mars2501 [29]
3 years ago
14

Which equations represent the line that is perpendicular to the line 5x – 2y = -6 and passes through the point

Mathematics
2 answers:
vitfil [10]3 years ago
6 0

Answer:

y+4=-(2/5)(x-5)

y=-(2/5)x-2

2x+5y=-10

Step-by-step explanation:

natta225 [31]3 years ago
4 0

Answer: 2x + 5y = - 10, Cy + 4 = (x-5)

Dy - 4 = (x+ 5)

Step-by-step explanation:

Equation of the line

5x - 2y = -6

Conditions for perpendicularity

m1 x m2 = -1

To get m1, rearrange the equation

2y = 5x + 6

y = 5x/2 + 3

n1 = 5/2 and m2 = -2/5

To get C

y = mx +c

-4 = -2 x 5/5 + C

-4 = -2 + C

C = -4 + 2

C = -2

To get the equation of the second line

y = -2x/5 - 2

Multiply through by 5

5y = -2x - 10

2x + 5y = 10.

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Solve the following initial-value problem:<br><br> (ex+y)dx+(3+x+yey)dy=0, y(0)=1<br><br> 0=
alex41 [277]

Answer:

e^x+xy+3y+(y-1)e^y=4

Step-by-step explanation:

Given that

(e^x+y)dx+(3+x+ye^y)dy=0

Here

M=e^x+y

N=3+x+ye^y

We know that

M dx + N dy=0 will be exact if

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

So

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it means that this is a exact equation.

\int d\left(e^x+xy+3y+(y-1)e^y\right)=0

Noe by integrating above equation

e^x+xy+3y+(y-1)e^y=C

Given that

x= 0 then y= 1

e^0+0+3+(1-1)e^1=C

C=4

So the our final equation will be

e^x+xy+3y+(y-1)e^y=4

3 0
3 years ago
The pair of variables x=5, y=7 is the solution to the equation ax–2y=1. Find the coefficient a.
Wewaii [24]

Answer:

a=3

Step-by-step explanation:

This should be very simple, just substitute xy and y for 5 and 7 respectively, so 5a-2*7=1, or 5a-14=1, or 5a=15, if we divide both sides by 5, we get a=3

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3 years ago
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Step-by-step explanation:

4 0
2 years ago
3+ 6x - x = 33<br> Can you find the answer
SVEN [57.7K]

Answer:

x=6

Step-by-step explanation:

6x-x=5x

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3 years ago
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