Rewrite without absolute value for the given conditions: y=|x−3|+|x+2|−|x−5|, if x<−2

y=-x-4 given that x is less than -2.

Step-by-step explanation:

|x-3|=x-3 when x-3 is positive, . So solving by adding 3 on both sides gives

|x-3|=-(x-3) when x-3 is negative, . So solving by adding 3 on both sides gives

is included in the last one since these values are less than 3. This means for our problem with given condition that |x-3|=-(x-3)=-x+3.

|x+2|=x+2 when x+2 is positive, . So solving by subtracting 2 on both sides gives

|x+2|=-(x+2) when x+2 is negative, . So solving by subtracting 2 on both sides gives

is included in the last one since these values are less than -2. This means for our problem with given condition that |x+2|=-(x+2)=-x-2.

|x-5|=x-5 when x-5 is positive, . So solving by adding 5 on both sides gives

|x-5|=-(x-5) when x-5 is negative, . So solving by adding 5 on both sides gives

is included in the last one since these values are less than 5. This means for our problem with given condition that |x-5|=-(x-5)=-x+5.

Lets put it altogether now.

y=|x−3|+|x+2|−|x−5|, if x<−2

y=-x+3+-x-2-(-x+5)

y=-x+3-x-2+x-5

y=-x-x+x+3-2-5

y=-x-4

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.