Factor the equation so...
(r^2-pr) and (p^2q-pqr)
Factor out (r^2-pr) = r(r-p)
Factor out (p^2q-pqr) = pq(p-r)
Add a negative to r(r-p) to make it -r(p-r)
(pq-r)(p-r) is the answer... I'm sorry I can't explain things well, but I tried.
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The two number are 39 and 13
<em><u>Solution:</u></em>
Let the two numbers be "a" and "b"
Let the larger number be "a" and the smaller number be "b"
<em><u>Given that, sum of two numbers is 52</u></em>
a + b = 52 ---------- eqn 1
<em><u>One number is 3 times as large as the other number</u></em>
Larger number = 3 times smaller number
a = 3b -------- eqn 2
<em><u>Let us solve eqn 1and eqn 2</u></em>
<em><u>Substitute eqn 2 in eqn 1</u></em>
3b + b = 52
4b = 52
b = 13
<em><u>Substitute b = 13 in eqn 2</u></em>
a = 3(13)
a = 39
Thus the two number are 39 and 13
Let AM be the distance between point A and the right wall and AN be the distance between A and the left wall.
Δ AMB is an isosceles right triangle and Δ ANC is half of an equilateral triangle.
Length of AM = 30 m. Length of AN = 1/2 · 80 = 40 m.
The distance between the walls is:
30 m + 40 m = 70 m.
You need the starting point and the end point and count how long it took to get to one point to the either
