6x-2x-9=7 solve for X there both the same sumber

Mathematics

1 answer:

mrs_skeptik [129]2 years ago

8 0

6x -2x -9= 7
⇒ 4x -9= 7
⇒ 4x= 7+ 9
⇒ 4x= 16
⇒ x= 16/4
⇒ x= 4

Final answer: x=4~

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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:

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

4 0

3 years ago

1. Rewrite the equation 5(x - 3) = x + 13 with 6 substituted for x. Write a

soldier1979 [14.2K]

Answer:

Step-by-step explanation:

5(6-3)=6+13 -simplify

30-15=6+13

15=19

x doesnt equal 6

5(x-3)=x+13

5x-15=x+13