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sp2606 [1]
3 years ago
10

Someone please explain this question PART "b) i" ONLY!!!

Mathematics
1 answer:
Gnom [1K]3 years ago
8 0
<h3>Answer:   (n-1)^2</h3>

This is because we have a list of perfect squares 0,1,4,9,...

We use n-1 in place of n because we're shifting things one spot to the left, since we start at 0 instead of 1.

In other words, if the answer was n^2, then the first term would be 1^2 = 1, the second term would be 2^2 = 4, and so on. But again, we started with 0^2 = 0, so that's why we need the n-1 shift.

You can confirm this is the case by plugging n = 1 into (n-1)^2 and you should find the result is 0^2 = 0. Similarly, if you tried n = 2, you should get 1^2 = 1, and so on. It appears you already wrote the answer when you wrote "Mark Scheme".

All of this only applies to sequence A.

side note: n is some positive whole number.

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5 0
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            Find the approximate solution of this system of equations.
Montano1993 [528]
Y = |x² - 3x + 1|
y = x - 1

|x² - 3x + 1| = x - 1
|x² - 3x + 1| = ±1(x - 1)
|x² - 3x + 1| = 1(x - 1)       or      |x² - 3x + 1| = -1(x - 1)
|x² - 3x + 1| = 1(x) - 1(1)    or    |x² - 3x + 1| = -1(x) + 1(1)
|x² - 3x + 1| = x - 1        or         |x² - 3x + 1| = -x + 1
  x² - 3x + 1 = x - 1         or          x² - 3x + 1 = -x + 1
        - x        - x                                + x         + x
  x² - 4x + 1 = -1           or            x² - 2x + 1 = 1
              + 1 + 1                                       - 1 - 1
  x² - 4x + 1 = 0              or           x² - 2x + 0 = 0
  x = -(-4) ± √((-4)² - 4(1)(1))    or    x = -(-2) ± √((-2)² - 4(1)(0))
                      2(1)                                             2(1)
  x = 4 ± √(16 - 4)            or            x = 2 ± √(4 - 0)
                 2                                                 2
  x = 4 ± √(12)              or               x = 2 ± √(4)
             2                                                  2
 x = 4 ± 2√(3)               or               x = 2 ± 2
             2                                                2
 x = 2 ± √(3)                or                x = 1 ± 1
 x = 2 + √(3)  or  x = 2 - √(3)   or    x = 1 + 1    or    x = 1 - 1
                                                      x = 2       or       x = 0
y = x - 1          or           y = x - 1                            or    y = x - 1   or    y = x - 1
y = (2 + √(3)) - 1    or    y = (2 - √(3)) - 1          or         y = 2 - 1    or    y = 0 - 1
y = 2 - 1 + √(3)     or      y = 2 - 1 - √(3)          or           y = 1      or       y = -1
y = 1 + √(3)        or        y = 1 - √(3)               (x, y) = (2, 1)    or    (x, y) = (0, -1)
       (x, y) = (2 ± √(3), 1 ± √(3))

The solution (0, -1) can be made by one function (y = x - 1) while the solution (2 ± √(3), 1 ± √(3)) can be made by another function (y = |x² - 3x + 1|). So the solution (2, 1) can be made by both functions, making the two solutions equal.
4 0
3 years ago
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