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Vaselesa [24]
3 years ago
11

HURRY PLZ!

Mathematics
1 answer:
timama [110]3 years ago
7 0

Answer:

D

Step-by-step explanation:

f^{-1}(x) means the inverse. The inverse is the function reflected over the y=x line. This means the values have been reversed. (x,y)-->(y,x). We are looking for a point that has reversed values from the table.

The table has (1.-5) so the inverse is (-5,1) or D.

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Answer:

3 37/90

Step-by-step explanation:

You didn't ask for the work. Hope this helps.

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4 0
3 years ago
What is cos q if sec q = 2? I need help please please
nirvana33 [79]
I hope this helps you


sec Q=1/cos Q


2=1/cos Q


cos Q= 1/2


Q=30+2.pi.n


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6 0
3 years ago
Hello i need help please for my testt!
blsea [12.9K]

Answer:

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Step-by-step explanation:

8 0
3 years ago
Write the equation in exponential form using the two points (0,1) and (1,3)
MakcuM [25]

Answer:

y=3^x

Step-by-step explanation:

The equation of the function in exponential form is

y=a\cdot b^x

The function is determined using points (0,1) and (1,3), so their coordinates satisfy the eduation. Substitute them:

1=a\cdot b^0\Rightarrow a=1\ \ [b^0=1]\\ \\3=a\cdot b^1\Rightarrow 1\cdot b=3,\ b=3

Thus, the equation of the function is

y=3^x

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
3 years ago
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