Answer:
In the given figure the point on segment PQ is twice as from P as from Q is. What is the point? Ans is (2,1).
Step-by-step explanation:
There is really no need to use any quadratics or roots.
( Consider the same problem on the plain number line first. )
How do you find the number between 2 and 5 which is twice as far from 2 as from 5?
You take their difference, which is 3. Now splitting this distance by ratio 2:1 means the first distance is two thirds, the second is one third, so we get
4=2+23(5−2)
It works completely the same with geometric points (using vector operations), just linear interpolation: Call the result R, then
R=P+23(Q−P)
so in your case we get
R=(0,−1)+23(3,3)=(2,1)
Why does this work for 2D-distances as well, even if there seem to be roots involved? Because vector length behaves linearly after all! (meaning |t⋅a⃗ |=t|a⃗ | for any positive scalar t)
Edit: We'll try to divide a distance s into parts a and b such that a is twice as long as b. So it's a=2b and we get
s=a+b=2b+b=3b
⇔b=13s⇒a=23s
The range is the y values so your answer is.
Range = 3, 7, 11
F(10) = (10)^2 + 1
= 100 + 1
= 101
101 is the answer
If two triangles are similar then the
corresponding sides are in proportion. Thus,
AB / AU = BC / UV = AC / AV
AB / (20x+108) = 703 / 444
Where AB is equivalent to:
AB = AU + UB
AB = 20x + 108 + 273
AB = 20x + 381
Therefore going back to the first equation:
(20x + 381) / (20x + 108)
= 703/444
444 (20x + 381) = 703 (20x + 108)
8880x + 169164 = 14060x + 75924
14060x - 8880x = 169164 – 75924
5180 x = 93240
x = 93240 / 5180
<span>x = 18 (ANSWER)</span>