REALY NEED HELP Find an equation of the line passing through the pair of points. Write the equation in the form Ax+By=C. (4,8) a
nd (3,5)
1 answer:
Let
<span>A (4,8) and B (3,5)
</span>step 1
find the slope m
m=(y2-y1)/(x2-x1)---------> m=(5-8)/(3-4)----> m=-3/-1----> m=3
step 2
with m=3 and point B (3,5) find the equation of the line
y-y1=m*(x-x1)-----> y-5=3*(x-3)----> y-5=3x-9----> 3x-y=4
the answer is
<span>the equation in the form Ax+By=C is
</span> 3x-y=4
see the attached figure
<span>A (4,8) and B (3,5)
</span>step 1
find the slope m
m=(y2-y1)/(x2-x1)---------> m=(5-8)/(3-4)----> m=-3/-1----> m=3
step 2
with m=3 and point B (3,5) find the equation of the line
y-y1=m*(x-x1)-----> y-5=3*(x-3)----> y-5=3x-9----> 3x-y=4
the answer is
<span>the equation in the form Ax+By=C is
</span> 3x-y=4
see the attached figure
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Answer:y=
Acosx+Bsinx +cosx ln(cosx)+x sinx
Step-by-step explanation:
given equation y''+y=secx
auxiliary equation
so CF is y=Acosx+Bsinx
now
using wronskian formula
=
now f(x)=secx
now particular integrals are
PI=cosx ln(cosx)+x sinx
total solution
y= C.F+P.I
y=Acosx+Bsinx +cosx ln(cosx)+x sinx
I found the correct image that accompanies this problem and edited it with my answers.. Pls. see attachment.
Based on the attachment, the correct statements are:
<span>1) DO,2 (x,y) = (2x, 2y)
2) Side Q'S' lies on a line with a slope of -1.
Q'(-6,6) S'(-2,2)
m = y1 - y2 / x1 - x2
m = 6 - 2 / -6 - (-2)
m = 4 / -4
m = -1
</span><span>5) The distance from Q' to the origin is twice the distance from Q to the origin.
</span>
Based on the attachment, the correct statements are:
<span>1) DO,2 (x,y) = (2x, 2y)
2) Side Q'S' lies on a line with a slope of -1.
Q'(-6,6) S'(-2,2)
m = y1 - y2 / x1 - x2
m = 6 - 2 / -6 - (-2)
m = 4 / -4
m = -1
</span><span>5) The distance from Q' to the origin is twice the distance from Q to the origin.
</span>