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serg [7]
3 years ago
14

What are the roots of the quadratic function f(b)=b^2-75

Mathematics
1 answer:
Ivahew [28]3 years ago
5 0
Set f(b) = b^2 - 75 = to 0 and solve for b:

b^2 = 75
b^2 = 25(3)
b = plus or minus sqrt(25[3]) = plus or minus sqrt(25)*sqrt(3) = plus or minus 5 sqrt(3).
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Mekhanik [1.2K]

Answer:

$1.81

Step-by-step explanation:

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8 0
3 years ago
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Factor of x^3-x^2-24x-36
Basile [38]
The trial and error method is used to find an initial factor:
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f(1) = (1)³ - (1)² - 24(1) - 36 = -60
f(2) = (2)³ - (2)² - 24(2) - 36 = -80
f(5) = (5)³ - (5)² - 24(5) - 36 = -56
*** f(6) = (6)³ - (6)² - 24(6) - 36 = 0 ***

f(6) = 0 so (x - 6) is a factor of f(x).
This means that: f(x) = x³ - x² - 24x - 36 = (x - 6)(ax² + bx + c).
To find a,b and c, use long division (or inspection) to divide x³ - x² - 24x - 36 by x - 6.
The other 2 factors of f(x) can then be found by factorizing the
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5 0
3 years ago
In one area, the cable company marked up the
butalik [34]

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3 years ago
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Otrada [13]

Answer:

6

Step-by-step explanation:

100/4 = 25%

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