An example for #1 would be that: 15 is divisible by 3, but not 9. every third multiple of 3, (9, 18, 27,...) is divisible by 9 because 9 is three times the size of 3.
For part two the first number that I thought of is 23.
<span>Simplifying
x2 + 8x + y2 + -2y = 64
Reorder the terms:
8x + x2 + -2y + y2 = 64
Solving
8x + x2 + -2y + y2 = 64
Solving for variable 'x'.
Reorder the terms:
-64 + 8x + x2 + -2y + y2 = 64 + -64
Combine like terms: 64 + -64 = 0
-64 + 8x + x2 + -2y + y2 = 0
The solution to this equation could not be determined.</span>
Answer:
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Step-by-step explanation:
12f+10<75
12f<75-10 *carry the 10 over
12f<65 *simplify
f<65/12 *carry the 12 over
f<5.42 *simplify
Let x be the 1st odd number, and x+2 the second odd consecutive number:
(x)(x + 2) = 6[((x) + (x+2)] -1
x² + 2x = 6(2x + 2) - 1
x² + 2x = 12x +12 - 1
And x² - 10x - 11=0
Solve this quadratic expression:
x' = [+10 +√(10²- 4.(1)(-11)]/2 and x" = [+10 -√(10²- 4.(1)(-11)]/2
x' = [10 + √144]/2 and x" = [10 - √64]/2
x' = (10+12)/2 and x" = (10-12)/2
x = 11 and x = -1
We have 2 solutions that satisfy the problem:
1st for x = 11, the numbers at 11 and 13
2nd for x = - 1 , the numbers are -1 and +1
If you plug each one in the original equation :(x)(x + 2) = 6[((x) + (x+2)] -1
you will find that both generates an equlity
