Ashley is packing her bags for her vacation. She has 7 unique socks, but only 3 fit in her bag.

Mathematics

1 answer:

IrinaVladis [17]4 years ago

5 0

Answer:

There are 35 possible group of 3 socks from a total of 7

Step-by-step explanation:

Given

Number of unique socks = 7

Number of possible selection = 3

Required

Number of ways she can make her selection.

The question can be categorised as combination; reason being that words like "take" as used in the question points to combination or selection.

To solve this, we make use of combination formula.

nCr = nPr/r!

Where n = 7 and r = 3

By substitution, we have

7C3 = 7P3÷3!

Taking it one at a time

7P3 = 7!÷(7-3)!

7P3 = 7! ÷ 4!

7P3 = 7 * 6 * 5 * 4! ÷ 4!

7P3 = 7 * 6 * 5

7P3 = 210

3! = 3 * 2 * 1

3! = 6.

So,

7C3 = 210 ÷ 6

7C3 = 35.

Hence, there are 35 possible group of 3 socks from a total of 7

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Determine all real numbers a$ such that the inequality $ |x^2 2ax 3a|\le2$ has exactly one solution in x$.

vovikov84 [41]

Answer:

$x=-1, S=\{x\:\in\mathbb{R}:x=-1\}$

Step-by-step explanation:

For the sake of clarity, assuming you meant:

$|x^{2}+2x+3|\le2$

1) Absolute Value or Modulus functions has one property that assures us that:

$If |a|$

2) Solving for x

$I) x^{2}+2x+3\geqslant -2\Rightarrow x^{2}+2x+5\geqslant 0 \Rightarrow x\geqslant \frac{-2\pm \sqrt{2^{2}-4*1*5}}{2} x'\geqslant \frac{-2+4i }{2} \:and\:x''\geqslant \frac{-2-4i }{2}$

Not defined in Real Set of Numbers

Evaluating $x^{2}+2x+3\leqslant 2$ :

$x^{2}+2x+3\leqslant 2\\x^{2}+2x+1\leqslant 0\\(x+1)(x+1)\leqslant 0 \Rightarrow S=\{-1\}$

3) So, since for the first case the Discriminant Δ <0, then the solutions presented for $x^{2}+2x+3\geqslant -2\: \in \mathbb{C}$ . The only solution in the Real Set for the inequality $|x^{2}+2x+3|\le2$ is $S=\{-1\}$ , i.e. x=-1.