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Korvikt [17]
3 years ago
6

Is it possible for a line to have a constant rate and not be proportional?

Mathematics
1 answer:
devlian [24]3 years ago
6 0
<h3>Answer: Yes it is possible</h3>

Explanation:

Any linear equation in the form y = mx+b has a constant rate of change, and that rate of change is the slope m. If b = 0, then it turns into y = mx. Here m is the constant of proportionality. Direct proportion equations always go through the origin. If b is nonzero, then y = mx+b will not be proportional.

So for example, y = 2x is proportional while y = 2x+1 is not proportional.

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Which of the following represents the zeros of f(x) = x3 − 11x2 + 38x − 40?
Delvig [45]
The answer is <span>5, 4, 2
</span>
Among all choices we have 5, so
x = 5
x - 5 = 0
Let's divide the expression by (x - 5) using the long division:
                          x³ - 11x² + 38x - 40
(x - 5)  * x² =      x³ - 5x²                   Subtract
____________________________
                              -6x² + 38x - 40
   (x - 5) * (-6x) =    -6x² + 30x          Subtract
____________________________
                                           8x - 40
                    (x - 5) * 8 =     8x - 40   Sutract
____________________________
                                              0

Thus: x³ - 11x² + 38x - 40 = (x - 5)(x² - 6x + 8)

Now, let's simplify x² - 6x + 8.

x² - 6x + 8 = x² - 2x - 4x + 8 =
                  = x² - 2*x - (4*x - 4*2) = 
                  = x(x - 2) - 4(x - 2) =
                  = (x - 4)(x - 2)

Hence:
x³ - 11x² + 38x - 40 = (x - 5)(x - 4)(x - 2)
To calculate zero:
x³ - 11x² + 38x - 40 = 0
(x - 5)(x - 4)(x - 2) = 0
x - 5 = 0            or             x - 4 = 0              or      x - 2 = 0
x = 5                 or              x = 4                   or      x = 2
3 0
3 years ago
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