Let 1st integer = xLet 2nd integer = x + 1 We set up an equation. x(x + 1) = 195 x2 + x = 195 x2 + x - 195 = 0
We will use the quadratic formula: x = (-b ± √(b2 - 4ac) / (2a) x = (-1 ± √(1 - 4(-195))) / 2 x = (-1 ± √(781)) / 2 x = (-1 ± 27.95) / 2 x = 13.48x = -14.78
<span>We determine which value of x when substituted gives us a product of 195.</span> 13.48(14.48) = 195.19-14.48(-13.48) = 195.19 <span>The solution is 2 sets of two consecutive number</span> <span>Set 1</span> The 1st consecutive integer is 13.48The 2nd consecutive integer is 14.48
<span>Set 2</span> The 1st consecutive integer is -14.48The 2nd consecutive integer is -13.48Hopefully this helped, hard work lol :)
<u>Answers:</u>
These are the three major and pure mathematical problems that are unsolved when it comes to large numbers.
The Kissing Number Problem: It is a sphere packing problem that includes spheres. Group spheres are packed in space or region has kissing numbers. The kissing numbers are the number of spheres touched by a sphere.
The Unknotting Problem: It the algorithmic recognition of the unknot that can be achieved from a knot. It defined the algorithm that can be used between the unknot and knot representation of a closely looped rope.
The Large Cardinal Project: it says that infinite sets come in different sizes and they are represented with Hebrew letter aleph. Also, these sets are named based on their sizes. Naming starts from small-0 and further, prefixed aleph before them. eg: aleph-zero.
The answer to your question is 11
Answer:
Step-by-step explanation:
61% = 0.61
0.68
2/3 = 0.666
0.57
3/5 = 0.600
least to greatest : 0.57, 3/5, 61%, 2/3, 0.68
I believe b but I could be so wrong I’m sorry if I am
