6<7-1/2x<7?

Answer 1

6<7-1/2x<7 | -7
-1<-1/2x<0 | *2
-2<-x<0 | *(-1)
2>x>0

x is in the interval (0;2)

Answer 2

6 < 7 -(1/2)x < 7
-13 < -(1/2)x < 0
0 < x < 26

6 < 7 -1/(2x) < 7
-13 < -1/(2x) < 0
0 < x < 2/13

Answer 3

build a triangle with factors of length a, b, c to correspond with the radians ?/7, 2?/7, and 4?/7 respectively. Bisect 4?/7 to 2?/7 and bisect between the radians back to ?/7. Now we are finished with construction what we %. it could be impossible to describe with the labeling, so label the vertex of radian ?/7 as A, the vertex of radian 2?/7 as B, and the vertex of 4?/7 as C. The bisector of radian C intersects the triangle at element D. the 2d bisector intersects are E. Triangle BCE is resembling triangle ABC, consequently | BE | = a million/c, | CE | = b/c. Triangle ACE is isosceles, subsequently | AE | = b. Triangle CDE is isosceles subsequently | CD | = b/c. Triangle CDE is resembling triangle ACE subsequently | DE | = b/c². provided that Triangle BCD is isosceles, b/c = a million/c + b/c². provided that Triangle ACE is isosceles, c = b + a million/c. b/c = a million/c + b/c² bc = b + c c = b + a million/c c² = bc + a million c² = b + c + a million c² - c - a million = b bc = b + c c³ - c² - c = c² - c - a million + c c³ - c² - c = c² - a million c³ - 2c² - c + a million = 0 resolve for c making use of the cubic formulation. After fixing for c, you are able to resolve for b. via using the regulation of Sines, sin (2?/7) and sin (4?/7) could nicely be expressed as sin (?/7). the only ultimate step is fixing for sin (?/7). sin (?/7) = (2 sin (?/7) cos (?/7)) / b b/2 = cos (?/7) b/2 = a million - sin² (?/7) sin (?/7) = ?(a million - b/2) Plug b in and resolve for it. nicely, you have sin (?/7) so which you in addition to could have the right fee.

Answer 4

subtract 7 from everything.

Multiply all terms by -2.

Switch the inequality signs.

Hopefully this isn't one of my students that posted this!

don't forget to graph.

Answer 5

the value of X

greater than zero - and less than 2

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