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vaieri [72.5K]
3 years ago
6

-4 X 4 X 4 X 4 X 4 Solve the problem

Mathematics
2 answers:
AveGali [126]3 years ago
7 0
The answer to this question I’m not sure of
Nuetrik [128]3 years ago
6 0

Answer:

-1,024

Step-by-step explanation:

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What is 22% of 150? i need this answer know!!!!
Scorpion4ik [409]

Answer:

33

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
I need help with #4 a-d
lesya [120]

a repeating decimal is placed over 9's instead of 0's.   for example: 0.23 = \frac{23}{100}, but 0.232323... = \frac{23}{99}

a) x = 0.121212

b) 100(x) = 100(0.121212)

    100x  = 12.1212

c)  100x = 12.1212

-  <u>        x =   0.1212</u>

     99x  =  12

         x  = \frac{12}{99}

d) -2\frac{12}{99}  <em>= -2\frac{4}{33} when simplified</em>

The number is rational becase it has a repeating decimal

6 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
K+ OH-<br> What’s is the formula for this
ruslelena [56]

Answer:

check the periodic table

3 0
2 years ago
A telephone pole has a wire to its top that is anchored to the ground. The distance from the bottom of the pole to the anchor po
Bumek [7]

Answer:

The height of the pole is 105ft,

Step-by-step explanation:

Let us call h the height of the pole, then we know that the distance from the bottom of the pole to the anchor point is 49, or it is h - 49.

The wire length d is 14 ft longer than height h, hence

d= h+14.

Thus we get a right triangle with hypotenuse d= y+14, perpendicular  h, and base h-49; therefore, the Pythagorean theorem gives

(h-49)^2+h^2 = (h+14)^2

which upon expanding we get:

h^2-98h+2401 = h^2+28h+196

further simplification gives

h^2-126h+2205=0,

which is a quadratic equation with solutions

h =21ft\\h = 105ft.

Since the first solution h =21ft will give the triangle base length of 21ft-49ft = -28ft which is negative; therefore, we disregard it and pick the solution h = 105ft.

Hence, the height of the pole is 105ft.

3 0
2 years ago
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