Give the number a label. That can be anything you want. A lot of people will use 'x' every single time they do a math problem,but there's no reason to do that and it's boring. Let's call our number ' M ' for 'Mystery number'. OK ?
The number . . . M
The square of the number . . . M²
Two more than the square of the number . . . M² + 2
You said that this is equal to 123, so we can write <u> M² + 2 = 123</u>
That's the equation we have to take and solve for ' M '.
Subtract 2 from each side of the equation, and you have M² = 121 .
Take the square root of each side: M = √121 .
The Mystery number is the square root of 121.
If you don't happen to know what that is, then you can use your pocket
calculator, or the calculator that comes with your computer (if you know
how to find it). They will all tell you that the square root of 121 is <em>11</em> .
That's a fine and wonderful answer, but technically, it's only half of the
answer. Any equation that has something squared in it almost always
has two solutions, and this one does.
The square root of 121 is a number that gives you 121 when you
multiply it by itself. ' 11 ' does that: (11 x 11) = 121 . Is there <em><u>another</u></em>
number that does the same thing ?
How about ' -11 ' ? Look at this: ( -11 x -11 ) = 121 . (Remember that
if both numbers being multiplied have the <em>same sign</em>, then their product
is positive.)
The bottom line is: The mystery number is<em> +11</em> and also<em> -11</em> .
Either one does what you want . . . When you square it and then
add 2 more, you get 123 either way.
Answer:y would equal 2 because when you put a symbol/number next to a number/variable you multiply the two #s/variables and 8x2 would equal 16
Step-by-step explanation:
Answer:
8/5 or 1 and 3/5
Step-by-step explanation:
substitute then add 4/5 and 4/5.
Answer:
Converges
Step-by-step explanation:
n = k+1
(k+1) ÷ [ln(k+1)]^(k+1)
n = k
k ÷ [ln(k)]^k
{(k+1) ÷ [ln(k+1)]^(k+1)} ÷ {k ÷ [ln(k)]^k}
= {(k+1)/k} × {[ln(k)]^k}/{[ln(k+1)]^(k+1)}
{[ln(k)]^k}/{[ln(k+1)]^(k+1)} is decreasing at a much faster rate than the rate of increase of {(k+1)/k}.
Therefore the product is less than 1.
Hence it converges
