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1 year ago

In the regular decagon shown, what is the measure of angle 1?

Mathematics

1 answer:

NeX [460]1 year ago

6 0

Answer:

try 36 but im not quite sure

Step-by-step explanation:

im sry if im wrong have a good rest of your day :)

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How to solve using Substitution or Elimination?<br> y=3x-1<br><br> y=2x+2

n200080 [17]

By elimination:

y = 3x - 1

y = 2x + 2

Subtract the second equation from the first

0 = x - 1

y = 2x + 2

Subtract the first equation from the second

0 = x - 1

y = x + 3

Subtract the first equation from the second again

0 = x - 1

y = 4

Subtract x from both sides of the first equation

- x = - 1

y = 4

Divide the first equation by (-1)

x = 1

y = 4

<h3>So, the solution is x = 1 and y = 4 {or: (1, 4)}</h3>

4 0

2 years ago

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.