Refer to the figure given below while reading the solution.
Suppose the dog reaches position A when traveled 10 m diagonally towards the opposite side.
And then position B when traveled 5 m towards the right turning 90°.
We can observe that APC is a right triangle with legs of equal length AC. And the coordinates of the point A is (AC, AC).
Also we can observe that APB is a right triangle with legs of equal length AD. Then the coordinates of the point D is (AC, AC-AD).
Hence, the coordinates of B will be (AC+AD, AC-AD).
Now, we since we have the coordinates we can calculate the shortest distances of B from each of the sides.
- The shortest distance of B from PQ = AC-AD
- The shortest distance of B from SR = 44-(AC-AD)
- The shortest distance of B from SP = AC+AD
- The shortest distance of B from RQ = 44-(AC+AD)
So, the average of the shortest distances of B from each side is 
Hence, the average of the shortest distance of B from each side is 22 m
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4%• x =56
x=56•100/4
x=5600/4
x=1400
Answer:
437.4m
Step-by-step explanation:
1 cm = 4.5m
7.2x4.5=32.4
3x4.5=13.5
32.4x13.5 = 437.4m
Degree is exponent, so 5th degree
difference of 2 perfect squares
a^2-b^2=(a-b)(a+b)
t^2-5^2=(t-5)(t+5)
2nd one
complete square
take 1/2 of linear coefient and square it
-6/2=-3, (-3)^2=9
add that to both sides
x^2-6x+9=-8+9
x^2-6x+9=1
factor
(x-3)^2=1
sqrt
take neg and pos root
x-3=+/-1
add 3
x=3+/-1
x=3+1 or x=3-1
x=4 or 2
3rd option