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Alchen [17]
3 years ago
12

For the sequence below, identify the type of sequence and its important features:

Mathematics
2 answers:
In-s [12.5K]3 years ago
4 0

Answer:

neither

Step-by-step explanation:

the sequence doesn't go up by a specific amount each time nor does it gets multiplied by a single number each time. there is no pattern

Verdich [7]3 years ago
4 0
Neither because it is going up at random
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David wants to hang a mirror in his room. The mirror and frame must have an area of 8 square feet. The mirror is 2 feet wide and
hoa [83]

Answer:

4x^2+10x-2=0

Step-by-step explanation:

So the total area of mirror: 2*3 = 6 ft sq

8-6 = 2 ft sq is the area of the frame

x is the thickness of the frame

so we have (2*x)*2+(3*x)*2+x*x*4= 2

4x+6x+4x^2=2

-> 4x^2+10x-2=0

8 0
2 years ago
Find the slope of the line passing through the points (-7,9) and (2,9).<br> slope=
Andru [333]

Answer:  The slope is 0

Step-by-step explanation:

The slope is the difference between the y coordinates divided by the difference between the x coordinates.

Y:  9 - 9 = 0

x: -7-2 =   -9

0/-9= 0  

7 0
3 years ago
Read 2 more answers
I need help
agasfer [191]
(0,4) hope this helps
7 0
3 years ago
Read 2 more answers
4x^2-24x+4y^2+72y=76
podryga [215]
<span>Simplifying 4x2 + -24x + 4y2 + 72y = 76
Reorder the terms: -24x + 4x2 + 72y + 4y2 = 76
Solving -24x + 4x2 + 72y + 4y2 = 76
Solving for variable 'x'.
Reorder the terms: -76 + -24x + 4x2 + 72y + 4y2 = 76 + -76
Combine like terms: 76 + -76 = 0 -76 + -24x + 4x2 + 72y + 4y2 = 0
 Factor out the Greatest Common Factor (GCF), '4'. 4(-19 + -6x + x2 + 18y + y2) = 0
 Ignore the factor 4.

</span><span>Subproblem 1
Set the factor '(-19 + -6x + x2 + 18y + y2)' equal to zero and attempt to solve: Simplifying -19 + -6x + x2 + 18y + y2 = 0 Solving -19 + -6x + x2 + 18y + y2 = 0 The solution to this equation could not be determined. This subproblem is being ignored because a solution could not be determined.
The solution to this equation could not be determined.</span>
7 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
2 years ago
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