Answer:
x^2 -6x + 222/25
Step-by-step explanation:
If the zeros are as above, then ;
x = 3-√3/5 or x = 3 + √3/5
Firstly, let’s represent √3/5 by b
Thus;
The two roots are ;
x = 3-b or x = 3 + b
so;
x+ b -3 and x -3-b
The quadratic equation is the product of the two
(x + b-3)(x - b -3)
x(x - b-3) + b(x -b -3) -3(x - b -3)
= x^2 -bx -3x + bx -b^2 -3b -3x + 3b + 9
Collect like terms and we are left with;
x^2 -6x -b^2 + 9
So let’s put back b = √3/5
x^2 -6x -(√3/5)^2 + 9
x^2 -6x -3/25 + 9
x^2 -6x + 222/25
Answer:g(x)=-4^2+4x-5
Step-by-step explanation:
Answer:
teacher got the answer physical theripists wrong on usatestprep
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Step-by-step explanation:
Answer:
Any [a,b] that does NOT include the x-value 3 in it.
Either an [a,b] entirely to the left of 3, or
an [a,b] entirely to the right of 3
Step-by-step explanation:
The intermediate value theorem requires for the function for which the intermediate value is calculated, to be continuous in a closed interval [a,b]. Therefore, for the graph of the function shown in your problem, the intermediate value theorem will apply as long as the interval [a,b] does NOT contain "3", which is the x-value where the function shows a discontinuity.
Then any [a,b] entirely to the left of 3 (that is any [a,b] where b < 3; or on the other hand any [a,b] completely to the right of 3 (that is any [a,b} where a > 3, will be fine for the intermediate value theorem to apply.
