Answer:
<h2>
64/27</h2>
Step-by-step explanation:
If x and y be real numbers satisfying 2/x=y/3=x/y, then any two of the equation are equated as shown;
2/x = y/3 ... 1 and;
y/3 = x/y... 2
From equation 1, 2y = 3x ... 3
and from equation 2; y² = 3x ... 4
Equating the left hand side of equation 3 and 4 since their right hand sides are equal, we will have;
2y = y²
2 = y
y = 2
Substituting y = 2 into equation 3 to get the value of x;
2y = 3x
2(2) = 3x
4 = 3x
x = 4/3
The value of x³ will be expressed as (4/3)³ = 4*4*4/3*3*3 = 64/27
The first and the third statements describe Jan's situation.
Answer:
-3>_ x >_ -2
Step-by-step explanation:
x(x+5)>_ -6
Open the bracket
x² + 5x >_ -6
x² +5x + 6 >_ 0
Solve quadratically using factorisation method
Factors of 6 that sum up to +5. ( 2 and 3)
x² + 2x + 3x + 6 >_ 0
Factorise
x(x + 2) +3(x + 2) >_ 0
(x+2)(x+3) >_ 0
Therefore
x + 2 >_ 0 or x +3 >_ 0
x>_ -2 or x >_ -3
-3>_ x >_ -2
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