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serg [7]
2 years ago
8

How many distinct products can be formed using two different integers from the given set: {–6, –5, –4, –3, –2, –1, 0, 1, 2, 3, 4

, 5}?
Mathematics
1 answer:
zhannawk [14.2K]2 years ago
6 0

Number of distinct products that can be formed is 144

<h3>Permutation</h3>

Since we need to multiply two different integers to be selected from the set which contains a total of 12 integers. This is a permutation problem since we require distinct integers.

Now, for the first integer to be selected for the product, since we have 12 integers, it is to be arranged in 1 way. So, the permutation is ¹²P₁ = 12

For the second integer, we also have 12 integers to choose from to be arranged in 1 way. So, the permutation is ¹²P₁  = 12.

<h3>Number of distinct products</h3>

So, the number of distinct products that can be formed from these two integers are ¹²P₁ × ¹²P₁ = 12 × 12 = 144

So, the number of distinct products that can be formed is 144

Learn more about permutation here:

brainly.com/question/25925367

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If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?
lys-0071 [83]

Answer:

2

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

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c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

3x^2-(3k-2)x-(k-6) \text{ with }k=2:

3x^2-(3\cdot 2-2)x-(2-6)

3x^2-4x+4

I'm going to solve 3x^2-4x+4=0 for x using the quadratic formula:

\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}

\frac{4\pm \sqrt{16-16(3)}}{6}

\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}

\frac{4\pm 4\sqrt{-2}}{6}

\frac{2\pm 2\sqrt{-2}}{3}

\frac{2\pm 2i\sqrt{2}}{3}

Let's see if uv=u+v holds.

uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}

Keep in mind you are multiplying conjugates:

uv=\frac{1}{9}(4-4i^2(2))

uv=\frac{1}{9}(4+4(2))

uv=\frac{12}{9}=\frac{4}{3}

Let's see what u+v is now:

u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}

u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}

We have confirmed uv=u+v for k=2.

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