The numbers DO differ by 10 if they are
- 1 and 11
- 2 and 12
- 3 and 13
- 4 and 14
- 5 and 15
Each outcome technically has two ways of occurring, but since we're taking two marbles at a time, that would be the same as saying, for instance, that drawing 1 and 11 is the same as drawing 11 and 1. We only count such an event once.
So the answer is (a) 5.
-6m+19
1/2(-12m+38)
All we have to do is simply distribute the 1/2.
1/2(-12m) + 1/2(38)
(-6m) + 19
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Arithmetic sequences have a common difference (addition)
geometric sequences have a common ratio (multiplication)
Answer:
x^4 -53x^2 +108x +160
Step-by-step explanation:
If <em>a</em> is a zero, then (<em>x-a</em>) is a factor. For the given zeros, the factors are ...
p(x) = (x +8)(x +1)(x -4)(x -5)
Multiplying these out gives the polynomial in standard form.
= (x^2 +9x +8)(x^2 -9x +20)
We note that these factors have a sum and difference with the same pair of values, x^2 and 9x. We can use the special form for the product of these to simplify our working out.
= (x^2 +9x)(x^2 -9x) +20(x^2 +9x) +8(x^2 -9x) +8(20)
= x^4 -81x^2 +20x^2 +180x +8x^2 -72x +160
p(x) = x^4 -53x^2 +108x +160
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The graph shows this polynomial has the required zeros.