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sladkih [1.3K]
3 years ago
12

Simplify 5/15 and 14/8

Mathematics
1 answer:
umka2103 [35]3 years ago
3 0

Answer:5/15 equals 1/3

Step-by-step explanation: i just simplified

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Rearrange e = 2f to make fthe subject.<br> =
GrogVix [38]

Answer:

f = e/2

Step-by-step explanation:

e = 2f

Switch sides.

2f = e

Divide both sides by 2.

f = e/2

6 0
3 years ago
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What's ∣x−2∣=∣4+x∣????
LuckyWell [14K]

Answer:

in the chapter 7 in class 9 Ex 7.3

one time -  and second time +

Step-by-step explanation:

x-2=4+x

1) x-2=-(4+x)

  x-2=-4-x

  x+x=-4+2

  2x=-2

  x=-2/2

  x=-1

2) x-2=4+x

   =0

7 0
3 years ago
The top of a round table with a diameter of 5.5 feet is painted white. how many square feet are painted
Firlakuza [10]
The formula for area of a circle is:
A=(pi)r^2
If diameter is 5.5, divide by 2 to find radius is 2.75. 2.75^2 is 7.5625 then multiplied by 3.14 (pi) is 23.746.
Depending on how you round, about 23.7 square feet were painted.
5 0
2 years ago
How does the value represented by the digit 4 in R compare to the value represented by the digit 4 in T?
timurjin [86]

Answer:

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Step-by-step explanation:

4 0
2 years ago
Suppose that f: R --&gt; 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
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