Hello Harrisonkaden13, the <span>square root of 10 plus the square root of 5 is,
</span>
<span>Answer:
</span>The principal, real, root of:
<span>^2√10</span><span> = 3.16227766 </span>All roots:
3.16227766 or <span>−3.16227766</span>
10 is not a perfect square.
The principal, real, root of:
^2<span>√5 </span><span>= 2.23606798</span>All roots:
2.23606798 or <span>−2.23606798
</span>5 is not a perfect square
We add them and we get 5.39834564 or -5.39834564.
I'll say the first integer is x. The next consecutive odd number would be x+2. If the sum of the odd integers is 236, the equation would be
x + (x + 2) = 236
solve for x
2x + 2 = 236
subtract 2 from each side of the equation
2x = 234
divide both sides by 2
x = 117
117 is the first odd integer. to find the other integer (x + 2), substitute 117 for x, and you have 117 + 2, which equals 119
The two consecutive odd integers that add up to 236 are 117 and 119.
They are both the same equations, the procedure to obtain x value will be equivalent.
