The numerical length of JK is 21 units.
Given that, point J is on line segment IK.
We need to determine the numerical length of JK.
In geometry, a line segment is a part of a line that is bounded by two distinctend points and contains every point on the line that is between its endpoints.
Given JK=x+6, IJ=9, and IK=2x.
Now, IK=JK+IJ
⇒2x=x+6+9
⇒2x=x+15
⇒x=15
Now, JK=x+6=21
Therefore, the numerical length of JK is 21 units.
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The distance of the line segment is solved and length of JK = 21 units.
Given data:
Since point J is on the line segment IK, the sum of lengths JK and IJ should be equal to the length of the whole line segment IK.
IJ + JK = IK
IJ = 9
IK = 2x
Now, substitute the values and solve for JK:
9 + JK = 2x
To find JK, we need to isolate it on one side of the equation. Subtract 9 from both sides:
JK = 2x - 9
Now, we are also given that JK = x + 6, so we can set these two expressions equal to each other:
2x - 9 = x + 6
Subtract x from both sides:
2x - x = 6 + 9
x = 15
Now, substitute the value of x back into the expression for JK:
JK = 2(15) - 9
JK = 30 - 9
JK = 21
Hence, the numerical length of JK is 21 units.
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Factor:
5x^2 + 28x + 32 =======> Solution: ======> ( x + 4 ) ( 5x + 8 )
Steps: Factor
5x^2 + 28x + 32
Break the expressions into groups:
( 5x^2 + 20x ) + ( 8x + 32 )
Factor Out 5x^2 From 5x^2 + 20x ======> 5x( x + 4 )
Factor Out the 8 From 8x + 32 ========> 8 ( x + 4 )
= 5x (x + 4 ) + 8 ( x + 4 )
Factor Out Common term:
x + 4
Answer =====> Factor of; ==> 5x^2 + 28x + 32 ( x + 4 ) ( 5x + 8 )
Hope that helps!!!! : )
Answer: 0.14
Step-by-step explanation:
1) Find the total #of seniors. To do this add up 25+5+5 = 35.
2)Since we know the number of seniors that work just do 5/35. This simplifies to 0.14 rounded to the nearest hundreth.