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3 years ago

-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:

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 --> 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