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Ede4ka [16]
2 years ago
6

Sarah is saving to buy a new phone. She needs $150, and she has already saved $63. Write an equation to model this situation. Le

t x represent the amount of money Sarah needs.
Mathematics
1 answer:
goblinko [34]2 years ago
7 0

Answer: 150-x=63$

Step-by-step explanation:

to solve most missing number equations using addition or subtraction its usually pretty simple

1. figure out the first number. the first number is usually the goal amount. so 150

2. were do i put x? in an equation 150-63 would equal the amount she needs. so when you subtract x from 150 you'd get 63. you already know the goal number 150 goes first and the equation equals 63 so x is going to be whats subtracted from 150

3. put it all together 150(goal)-x(unknown number)=63(the amount she already had)

4.solve 150-87=63

5.check your work 150-63=87

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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

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