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Mariana [72]
3 years ago
10

the moon is about 240,000 miles from Earth. What is this distance written as a whole number multiplied by a power of ten?

Mathematics
1 answer:
Ronch [10]3 years ago
3 0
I'm not sure does anyone else know the answer?
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8 avocados cost $16.what is the cost of the one avocado
ivann1987 [24]

Divide 16 by 8.

16 ÷ 8 = 2.

So, one avocado costs $2.

8 0
3 years ago
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Data: (12 ,8 , 6 , 12 , 16 , 12 ) median ............. mode......
Vesna [10]
Ascending order=  6, 8, 12, 12, 12, 16

No of observations (n) = 6
Median = Mean of the values of (n/2) and (n/2+1)
             = Mean values of 3rd and 4th values
             = 12+12/2
             = 12
Therefore, median = 12

Mode = Most number of observations= 12
4 0
4 years ago
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6y²×3³×18² what do I do
koban [17]

Answer:

52488y²

Step-by-step explanation:

first solve for 3³. then solve for 18²

3³=27

18²=324

27*324=8748

then multiply 8748 by 6y²

=52488y^2

6 0
3 years ago
YOU WILL GET MARKED BRAINLIEST:)
Aleks04 [339]

Answer:

Z = 2

Step-by-step explanation:

This tape diagram is split into two equal parts.

Since 2 and 2 are equal, and since 2 and 2 have a sum of 4, then z is equal to 2.

7 0
3 years ago
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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

4 0
3 years ago
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