Point J is on line segment \overline{IK} IK . Given IK=5x, IK=5x, IJ=4,IJ=4, and JK=4x, JK=4x, determine the numerical length of \overline{IK}. IK

IK = 20

Step-by-step explanation:

If Point J is on the line segment IK, then;

IK = IJ + JK

Given that

IJ = 4

IK = 5x

JK = 4x

Substitute into the expression

5x = 4 + 4x

5x - 4x  = 4

x = 4

Get the value of IK

IK = 5x

IK = 5(4)

IK = 20

Hence the numerical length of IK is 20

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

a. she will receive a refund of $278.

step-by-step explanation: