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3 years ago

What is the relationship between DG and FG? Math

Mathematics

1 answer:

Feliz [49]3 years ago

4 0

Ans: DG is perpendicular to FG. OC DG Intersects FG to form an acute angle. ODDO intersects FG to form an obtuse angle.

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Triangle CAT was rotated to create triangle C’A’T’. Describe the transformation using details and degrees

Elis [28]

Answer:

As this triangle is being transformed by rotation, the angles will stay same as well as the sides and other details, if you are doing this on a coordinate plane, then the points of the triangle would be changed and nothing else

If my answer helped, please mark me as the brainliest!!

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3 years ago

What is the y-intercept?

tankabanditka [31]

The y-intercept is 10.

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3 years ago

Read 2 more answers

4)^-5 write in a simplified fraction

erastova [34]

$\bf 4^{-5}?\qquad \textit{well, bear in mind that}\\
----------------------------\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}\qquad \qquad

\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\\ \quad \\
% negative exponential denominator
a^{{ n}} \implies \cfrac{1}{a^{- n}}
\qquad \qquad 

\cfrac{1}{a^{- n}}\implies \cfrac{1}{\frac{1}{a^{ n}}}\implies a^{{ n}}\\
----------------------------\\
thus\implies 4^{-5}\implies \cfrac{1}{4^{+5}}\implies \cfrac{1}{4^5}\implies \cfrac{1}{1024}$

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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.

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3 years ago

Evaluate f(x) = x + 6 when x = -2,0 and 5

Margaret [11]

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3 years ago