Answer:
x = 12, sides = 23
Step-by-step explanation:
Since 
is equilateral, all three sides are equal. Therefore:
![\left[\begin{array}{l}y=2x-1\\y=5x-37\\y=x+11\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bl%7Dy%3D2x-1%5C%5Cy%3D5x-37%5C%5Cy%3Dx%2B11%5Cend%7Barray%7D%5Cright%5D)
Substituting y = x + 11:
![\left[\begin{array}{l}x+11=2x-1\\x+11=5x-37\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bl%7Dx%2B11%3D2x-1%5C%5Cx%2B11%3D5x-37%5Cend%7Barray%7D%5Cright%5D)
Solving for x:
Substituting x = 12:
Thus, <em>x</em> = 12 and the length of each side is 23.
The perfect squares on either side of 21 are 16 and 25 of which the roots are:
4 and 5
So, The dots represent the children.
And the numbers represent the length in.
I wouldnt say its D.
I would say its C. Group B, because 4 kids have forearm length longer then 13 in.
A) Probability that at least one of these golfers makes a hole in one on the 10th hole at the same tournament is
<span>=1−(<span>2493/2494</span><span>)^<span>5
=1-(0.99)^5
= 1- 0.96
=0.4
b) T</span></span></span><span>he probability that all of these golfers make a hole-in-one on the 10th hole at the same tournament
=</span>(<span>1/2494</span><span>)<span>5
</span></span>
which is nearly zero.
The coordinates of C' are 3 times those of C.
3×(2, -1) = (6, -3)
The appropriate selection is ...
(6, -3) . . . . . . . . . the 4th one
