PLEASE HELP ME WITH THIS PROBLEM At the elementary school, the fourth graders Jamal and Dinah are still eating lunch as they come into math class. Someone has left a book on the floor and they both trip. As they each hit the floor, the food they are carrying flies across the room directly toward Harold, who is showing off his latest dance moves.

As Jamal and Dinah watch in horror, Jamal’s cupcake and Dinah’s sandwich splat right on top of Harold’s head! Jamal’s cupcake flies on a path that would have landed on the floor 22 feet away from him if it had not hit Harold. Dinah’s sandwich flies on a path that would have landed on the floor 26 feet away from her if it had not hit Harold. Jamal’s cupcake flies 8 feet high, while Dinah’s sandwich reaches a height of 5 feet before hitting Harold.

How tall is Harold? Once you have the equations, you may use Desmos to help find the answer.

This question uses a numeric answer. Type in just the number part answer (no letters!) and round to 2 decimals.

its c

the answers are

1. 2

2. 5

3. symmetrical

step-by-step explanation:

the line y=x is the line where the x and y coordinates of each point, are equal.

so this line is the line through the origin (0, 0), which makes an angle of 45° with the x-axis.

we draw triangle abc. to reflect this triangle with respect to the line y=x, we draw perpendiculars from the points a, b and c to to line y=x, and extend the perpendiculars so that the lint y=x is a perpendicular bisector of these line segments. 

we see that a(5, 3) is mapped to a'(3, 5)

and all the other points p(a, b) of the original triangle are mapped to p'(b,a)

the last statement is true for each point x and its reflection x', also it is true that the line y=x is the perpendicular bisector of xx' and yy'.

to prove these, check the second picture.

let the coordinates of x be (a, b), we want to find the coordinates of its reflection point with respect to the line y=x, point x'.

let m be the point on the line y=x, such that xm=mx'

let n be the point (b, b) which surely is on the line.

now, x'm=mx, nm is common and m(x'mn)=m(xmn)=90°, so by side-angle-side congruence postulate, the triangles x'mn and xmn

  are congruent, so nx=nx'

clearly the x coordinate of x' is b, so the y coordinate of x' is 

b+nx'=b+nx=a

so the coordinates of x' are (b, a)

read more on -