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weqwewe [10]
3 years ago
14

What is used to locate a point on the coordinate plane?

Mathematics
1 answer:
DiKsa [7]3 years ago
5 0
You can locate<span> any </span>point on the coordinate plane<span> using an ordered pair of numbers. 

e.i. (4,2) will be 4 units to the right and 2 units up on the coordinate plane.</span>
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What is 20% of 95? (show all work)
elena-s [515]
<span>we know that if 95 is out of 100%, we can write it down as 95 = 100%. w</span><span>e also know that x is 20% of the output value, so we can write it down as x = 20%. </span>
now we have the equations:
95=100%
x=20%
next we will put the equations together
95/x = 100%/20%
95/x=100/20
<span>(95/x)*x=(100/20)*x      
</span>we multiply both sides of the equation by x
<span>95=5*x      
</span>now we divide both sides of the equation by 5 to get x
<span>95/5=x </span>
<span>19=x </span>
x=19
we now know that <span>20% of 95 = 19
let me know if you have any other questions
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6 0
3 years ago
the sum of the measures of two interior angles of a triangle is 593°. Find the measure of the sixth interior angle
soldi70 [24.7K]
The sixth angle would measure 127°

5 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
Look at the rectangle and the square:
VashaNatasha [74]

Answer:

She is incorrect

Step-by-step explanation:

Pythagorean theorem:

16^2+8^2

256+64=320

320 is 17.89

8^2+8^2

64+64=128

√128 is 11.31

Ada is incorrect. The length of diagonal SQ is bigger than the length of diagonal OM. But it is not two times bigger.

5 0
3 years ago
Which polynomial equation of least degree has -2, -2, 3, and 3 as four of its roots?
I am Lyosha [343]

Answer:

The and is x-2)-2(x + 3)3=0

4 0
3 years ago
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