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3 years ago

Need help on this question i dont get it ima be posting some more but i need help on this on first

Mathematics

1 answer:

Anuta_ua [19.1K]3 years ago

8 0

Answer:

The answer is B.

Step-by-step explanation:

Its B because in the problem it says $12 per yard which means it would be 12y. The 40 Rupesh already had so just add it and it would be 12y+ 40. The key word "At least" results into the sign being greater than or equal to. So the final answer would be 12y+40 $\geq$ 148.

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In the diagram of the trapezoid RSUT, RS is parallel to TU, X is the midpoint of RT, and V is the midpoint of SU.

grigory [225]

It’s 58 because the length is increasing by 14

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The perimeter of a rectangle is 52. The length of the rectangle is six less than the width. Find the length and width of the rec

devlian [24]

Create expressions for the length and width of the rectangle.

length (l): x - 6
width (w): x

Perimeter: 2l + 2w

Substitute (x - 6) for l and (x) for w.

52 = 2(x - 6) + 2(x)

Distribute 2 inside the parentheses.

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3 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.