Find the slope-intercept form of the

Mathematics

1 answer:

3241004551 [841]3 years ago

3 0

Parallel lines have the same slope.
So I plug in the -5 and -4
-4 = 1/2(-5) + b
-4 = -2.5 + b
b = -1.5

So I get y = 1/2x - 3/2

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Let c be the curve which is the union of two line segments, the first going from (0, 0) to (4, 4) and the second going from (4,

victus00 [196]

First of all we need to find a representation of C, so this is shown in the figure below.

So the integral we need to compute is this:

$I=\int_c 4dy-4dx$

So, as shown in the figure, C = C1 + C2, so:

$I=\int_{c_{1}} (4dy-4dx)+\int_{c_{2}} (4dy-4dx)=I_{1}+I_{2}$

Computing first integral:

$c_{1}: y-y_{0}=m(x-x_{0}) \rightarrow y=x$

Applying derivative:

$dy=dx$

Substituting this value into $I_{1}$

$I_{1}=\int_{c_{1}} (4dx-4dx)=\int_{c_{1}} 0 \rightarrow \boxed{I_{1}=0}$

Computing second integral:

$c_{2}: y-y_{0}=m(x-x_{0}) \rightarrow y-0=-(x-8) \rightarrow y=-x+8$

Applying derivative:

$dy=-dx$

Substituting this differential into $I_{2}$

$I_{2}=\int_{c_{2}} 4(-dx)-4dx=\int_{c_{2}} -8dx=-8\int_{c_{2}}dx$

We need to know the limits of our integral, so given that the variable we are using in this integral is x, then the limits are the x coordinates of the extreme points of the straight line C2, so:

$I_{2}= -8\int_{4}^{8}}dx=-8[x]\right|_4 ^{8}=-8(8-4) \rightarrow \boxed{I_{2}=-32}$

Finally:

$I=\int_c 4dy-4dx=0-32 \rightarrow \boxed{I=-32}$

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3 years ago

100 PTS FOR 1 QUESTION!!!!!!!

TiliK225 [7]

Answer:

9.8

Step-by-step explanation:

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