Mathematics: Point J is on line segment IK. Given JK 2x – 1, IK = 3x + 2, and IJ = 3x – 5, determine the numerical length of JK.

Answer: the correct answer is 10step-by-step explanation: let the number be n, then the expression becomes,then“twelve more than the quotient of a number and two, decreased by four” is[tex](rac{n}{2}+12)-4[/tex]when n=4,we substitute 4 whetever we see n to get,[tex](rac{4}{2}+12)-4[/tex]we simplify to obtain,[tex]=(2+12)-4[/tex]we further simplify to get,[tex]=14-4=10[/tex]therefore the correct answer is c....

Point J is on line segment IK. Given JK 2x – 1,

Ответ:

smortandsons

02.02.2022

Mathematics

JK = 7

Step-by-step explanation:

Point J is on line segment IK.

$\therefore JK + IJ = IK\\\therefore 2x-1 + 3x - 5 = 3x + 2\\\therefore 5x - 6 = 3x + 2\\\therefore 5x - 3x = 2 + 6\\\therefore 2x = 8 \\ \therefore x = \frac{8}{2} \\ \therefore x = 4 \\ \\ JK = 2x - 1 = 2 \times 4 - 1 = 8 - 1 \\ \huge \red{ \boxed{JK = 7}}$

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Ответ:

flex23

19.12.2022

Mathematics

Answer: the correct answer is 10step-by-step explanation: let the number be n, then the expression becomes,then“twelve more than the quotient of a number and two, decreased by four” is $(\frac{n}{2}+12)-4$ when n=4,we substitute 4 whetever we see n to get, $(\frac{4}{2}+12)-4$ we simplify to obtain, $=(2+12)-4$ we further simplify to get, $=14-4=10$ therefore the correct answer is c.

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Point J is on line segment IK. Given JK 2x – 1, IK = 3x + 2, and IJ = 3x – 5, determine the numerical length of JK.

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