Ok, I'm going to start off saying there is probably an easier way of doing this that's right in front of my face, but I can't see it so I'm going to use Heron's formula, which is A=√[s(s-a)(s-b)(s-c)] where A is the area, s is the semiperimeter (half of the perimeter), and a, b, and c are the side lengths.
Substitute the known values into the formula:
x√10=√{[(x+x+1+2x-1)/2][({x+x+1+2x-1}/2)-x][({x+x+1+2x-1}/2)-(x+1)][({x+x+1+2x-1}/2)-(2x-1)]}
Simplify:
<span>x√10=√{[4x/2][(4x/2)-x][(4x/2)-(x+1)][(4x/2)-(2x-1)]}</span>
<span>x√10=√[2x(2x-x)(2x-x-1)(2x-2x+1)]</span>
<span>x√10=√[2x(x)(x-1)(1)]</span>
<span>x√10=√[2x²(x-1)]</span>
<span>x√10=√(2x³-2x²)</span>
<span>10x²=2x³-2x²</span>
<span>2x³-12x²=0</span>
<span>2x²(x-6)=0</span>
<span>2x²=0 or x-6=0</span>
<span>x=0 or x=6</span>
<span>Therefore, x=6 (you can't have a length of 0).</span>
7(a - 10) = 13 - 2(2a + 3)
7a - 70 = 13 - 4a - 6 = 7 - 4a
7a + 4a = 7 + 70
11a = 77
a = 77/11 = 7
a = 7.
Answer:
(y times 6) -1
Step-by-step explanation:
I don't know what to say it is already simplified
Answer:
25.05
Step-by-step explanation:
Given that a random sample of 160 car crashes are selected and categorized by age. The results are listed below. The age distribution of drivers for the given categories is 18% for the under 26 group, 39% for the 26-45 group, 31% for the 45-65 group, and 12% for the group over 65.
<26 26-45 46-65` >65
Observed O 66 39 25 30 160
Expected E 40 40 40 40 160
Chi square
(O-E)^2/E 16.9 0.025 5.625 2.5 25.05
Thus we get chi square test statistic = 25.05
