HI THERE LET ME HELP U WITH YOUR QUESTION
**PERFECT SQUARES ARE POSITIVE INTEGERS**<span>
</span>THE NUMBERS CANNOT INCLUDE ITSELF AND ALWAYS SHOULD HAVE A ONE IN IT.
2.89 AND 0.004 ARE PERFECT NUMBERS. THEY INCLUDE 1 AND THEY DONT INCLUDE ITSELF
HOPE IT HELPS
Answer:
x : y = 9 : 13
Step-by-step explanation:
Given that
3x - y : x + 2y = 2 : 5
Express the ratio in fractional form, that is
= 
( cross- multiply )
5(3x - y) = 2(x + 2y) ← distribute parenthesis on both sides )
15x - 5y = 2x + 4y ( subtract 2x from both sides )
13x - 5y = 4y ( add 5y to both sides )
13x = 9y ( divide both sides by 13 )
x = 
y ( divide both sides by y )
= , that is
x : y = 9 : 13
The estimate of the problem would be 1/4 because 6/8- 1/2 you would get 1/4.
~Good Luck!~
If <em>x</em>² + <em>y</em>² = 1, then <em>y</em> = ±√(1 - <em>x</em>²).
Let <em>f(x)</em> = |<em>x</em>| + |±√(1 - <em>x</em>²)| = |<em>x</em>| + √(1 - <em>x</em>²).
If <em>x</em> < 0, we have |<em>x</em>| = -<em>x</em> ; otherwise, if <em>x</em> ≥ 0, then |<em>x</em>| = <em>x</em>.
• Case 1: suppose <em>x</em> < 0. Then
<em>f(x)</em> = -<em>x</em> + √(1 - <em>x</em>²)
<em>f'(x)</em> = -1 - <em>x</em>/√(1 - <em>x</em>²) = 0 → <em>x</em> = -1/√2 → <em>y</em> = ±1/√2
• Case 2: suppose <em>x</em> ≥ 0. Then
<em>f(x)</em> = <em>x</em> + √(1 - <em>x</em>²)
<em>f'(x)</em> = 1 - <em>x</em>/√(1 - <em>x</em>²) = 0 → <em>x</em> = 1/√2 → <em>y</em> = ±1/√2
In either case, |<em>x</em>| = |<em>y</em>| = 1/√2, so the maximum value of their sum is 2/√2 = √2.
