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3 years ago

A coat that cost $123 is marked 1/3 off what is the sale price of the coat

Mathematics

2 answers:

myrzilka [38]3 years ago

8 0

The answer would be 82$

alina1380 [7]3 years ago

4 0

I/3 of 123 is 41
41 dollars are taken off, so 123-41=82
the sale price is 82 dollars

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How many words of five letters can be created that start with the letter T, contain 2 other consonats and end with 2 vowels? Not

Jet001 [13]

If we're only counting 5 vowels (A, E, I, O, U) and 20 consonants (everything else, minus T), then there are

$\dbinom52=\dfrac{5!}{2!(5-2)!}=10$

ways of picking the vowels, and

$\dbinom{20}2=\dfrac{20!}{2!(20-2)!}=190$

ways of picking the consonants.

We want the word to start with T, and we'll allow any arrangement of the other 4 letters, so that the total number of words is

$4!\dbinom52\dbinom{20}2=\boxed{45,600}$

Keep in mind that this means words like TRIES and TIRES are treated as different.

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3 years ago

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RideAnS [48]

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Suppose that f : [a, b] → [a, b] is continuous. Prove that f has a fixed point. That is, prove that there exists c ∈ [a, b] such

nikklg [1K]

Answer:

Step-by-step explanation:

define the function:

$g(x) = f(x) -x$

As both $f(x)$ and x are continuous functions, $g(x)$ will also be continuous.

Now, what can we say about $g(a) = f(a) -a$ ?

we know that $a\leq f(a) \leq b$ , thus:

$a-a\leq f(a)-a \leq b-a\\0 \leq g(a) \leq b-a$

thus $g(a)$ is non-negative.

What about $g(b)$ ? Again we have:

$a\leq f(b) \leq b\\a-b \leq f(b) -b \leq 0\\a-b \leq g(b) \leq 0$

That means that $g(b)$ is not positive.

Now, we can imagine two cases, either one of $g(a)$ or $g(b)$ is equal to zero, or none of them is. If either of them is equal to zero, we have found a fixed point! In fact, any point $c$ for which $g(c)=0$ is a fixed point, because:

$g(c) = 0 \implies f(c) -c = 0 \implies f(c) = c$

Now, if $g(a) \neq 0$ and $g(b) \neq 0$ , then we have that

$g(a) >0$ and $g(b) < 0$ . And by Bolzano's theorem we can assert that there must exist a point c between a and b for which $g(c)=0$ . And as we have shown before that point would be a fixed point. This completes the proof.