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disa [49]
2 years ago
14

20/28 in its simplest form what is it ?

Mathematics
1 answer:
Sindrei [870]2 years ago
5 0
The answer is 5/7.

Divide 20/28 by the greatest common factor, 4

You will get 20/28 =5/7
You might be interested in
Solve for x. Write both solutions, separated by a<br> comma.<br> 9x^2+ 3x – 2 = 0
kogti [31]

Answer:

x=1/3, x=-2/3

Step-by-step explanation:

9x^2 +6x -3x -2=0

6x and -3x add up to 3x. You need four terms to factor.

Then you factor=

3x(3x+2)-1(3x+2)=0

(3x-1)(3x+2)=0

Then you can make two equations

3x-1=0 and 3x+2=0 because to multiply to get 0, at least one will be 0.

3x-1=0

3x=1

x=1/3

3x+2=0

3x=-2

x=-2/3

So your answers are x=1/3, x=-2/3

hope this helps!

6 0
3 years ago
A 5-month membership to the gym costs $125 jim would like to be a member for 8 months. What is the total amount he Will pay for
SCORPION-xisa [38]

Answer:

$200

Step-by-step explanation:

125/5=25

25 x 8=200

5 0
3 years ago
Math- u can zoom in to see the numbers on the chart and u can just tell me we’re the but the dots at and the awnsers pls
Misha Larkins [42]

Answer:

hgjajshdhjdjdjjdjfjjdjj

4 0
3 years ago
i a man even number beetween 50 and 99 i am a multiple of 9 my tens digit is 5 more than my units digit
vodomira [7]
Hello,
multiple of 9 betheen 50 and 99 are :54,63,72,81,99
and 54,72  are even.
5-4=1 and not 5
7-2=5 ok

72 is the number searched.
6 0
3 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
3 years ago
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