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bonufazy [111]
3 years ago
12

Proofs with transformations

Mathematics
1 answer:
nekit [7.7K]3 years ago
4 0

I don't really understand the question or else I would help

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Can someone please help me with this?
fomenos

perimeter = Answer:

Step-by-step explanation:

5. Perimeter of semi circle = r(pi+2)

Given diameter=17m; r=17/2

Perimeter = (17/2)(pi+2)

6. Given diameter = 10.5

Perimeter = (10.5/2)(pi+2)

8 0
3 years ago
NEED HELP NOW 71 POINTS WILL MARK BRAINLIEST ANSWER!!!!!!
andreyandreev [35.5K]
The answers are Rotation and Reflection: reflect horizontally and the V will be flipped upside down, rotate 180 degrees and the V will also be flipped.
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3 years ago
Read 2 more answers
Which graph represents a function
Oduvanchick [21]

Answer:

A

Step-by-step explanation:

Linear as said in it's name is a function that creates (visualy) a straight line.

FROM all the graphs A shows a straight line therefore, the answer.

8 0
3 years ago
Suppose that f: R --> R is a continuous function such that f(x +y) = f(x)+ f(y) for all x, yER Prove that there exists KeR su
Pachacha [2.7K]
<h2>Answer with explanation:</h2>

It is given that:

f: R → R is a continuous function such that:

f(x+y)=f(x)+f(y)------(1)  ∀  x,y ∈ R

Now, let us assume f(1)=k

Also,

  • f(0)=0

(  Since,

f(0)=f(0+0)

i.e.

f(0)=f(0)+f(0)

By using property (1)

Also,

f(0)=2f(0)

i.e.

2f(0)-f(0)=0

i.e.

f(0)=0  )

Also,

  • f(2)=f(1+1)

i.e.

f(2)=f(1)+f(1)         ( By using property (1) )

i.e.

f(2)=2f(1)

i.e.

f(2)=2k

  • Similarly for any m ∈ N

f(m)=f(1+1+1+...+1)

i.e.

f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)

i.e.

f(m)=mf(1)

i.e.

f(m)=mk

Now,

f(1)=f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=f(\dfrac{1}{n})+f(\dfrac{1}{n})+....+f(\dfrac{1}{n})\\\\\\i.e.\\\\\\f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=nf(\dfrac{1}{n})=f(1)=k\\\\\\i.e.\\\\\\f(\dfrac{1}{n})=k\cdot \dfrac{1}{n}

Also,

  • when x∈ Q

i.e.  x=\dfrac{p}{q}

Then,

f(\dfrac{p}{q})=f(\dfrac{1}{q})+f(\dfrac{1}{q})+.....+f(\dfrac{1}{q})=pf(\dfrac{1}{q})\\\\i.e.\\\\f(\dfrac{p}{q})=p\dfrac{k}{q}\\\\i.e.\\\\f(\dfrac{p}{q})=k\dfrac{p}{q}\\\\i.e.\\\\f(x)=kx\ for\ all\ x\ belongs\ to\ Q

(

Now, as we know that:

Q is dense in R.

so Э x∈ Q' such that Э a seq belonging to Q such that:

\to x )

Now, we know that: Q'=R

This means that:

Э α ∈ R

such that Э sequence a_n such that:

a_n\ belongs\ to\ Q

and

a_n\to \alpha

f(a_n)=ka_n

( since a_n belongs to Q )

Let f is continuous at x=α

This means that:

f(a_n)\to f(\alpha)\\\\i.e.\\\\k\cdot a_n\to f(\alpha)\\\\Also\\\\k\cdot a_n\to k\alpha

This means that:

f(\alpha)=k\alpha

                       This means that:

                    f(x)=kx for every x∈ R

4 0
2 years ago
If x = −3, which number line shows the value of |x|? (5 points)
Arada [10]
Can’t see your answer choices but the absolute value of -3 is 3 so I would say 3 is your answer
3 0
2 years ago
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