With digits 8, 5, and 6, an odd number must end in 5.
So far, we have _ _ 5.
The 8 is in the hundreds place.
Now we have 8_5.
The only place left for the 6 is the tens place.
The number is 865.
Answer:
Answer=C
Step-by-step explanation:
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 :)
Is the number that satys the same the number that changes the number that conrtolles them
