<h2>
Answer with explanation:</h2>
It is given that:
f: R → R is a continuous function such that:
∀ x,y ∈ R
Now, let us assume f(1)=k
Also,
( 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,
i.e.
f(2)=f(1)+f(1) ( By using property (1) )
i.e.
f(2)=2f(1)
i.e.
f(2)=2k
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,

Also,
i.e. 
Then,

(
Now, as we know that:
Q is dense in R.
so Э x∈ Q' such that Э a seq
belonging to Q such that:
)
Now, we know that: Q'=R
This means that:
Э α ∈ R
such that Э sequence
such that:

and


( since
belongs to Q )
Let f is continuous at x=α
This means that:

This means that:

This means that:
f(x)=kx for every x∈ R
Answer:
10
Step-by-step explanation:
Let the number of blue balls be x.
Number of red balls = 5
Total number of balls = x + 5
P (getting a red ball) = 5/x+5
P (getting a blue ball) = x/x+5
Given that
2(5/x+5) = x/x+5
10(x+5) = x^(2) +5x
x^(2)-5x-50=0
x(x-10)(x+5)
Now either x-10 = 0 or x+5= 0
But number of balls cannot be negative.
Hence, the number of blue balls is 10.
x^2=4x-5
subtract 4x from both sides
x^2-4x=-5
add 5 to both sides
x^2-4x+5=0
input into quadratic formula which is x=

or

si ax^2+bx+c
so a=1
b=-4
c=5
input

=

=

=
![\frac{4+2 times \sqrt{-1} }{2}= \frac{6 times \sqrt{-1} }{2}=3 times \sqrt{-1} [\tex][\tex]\sqrt{-1}](https://tex.z-dn.net/?f=%20%5Cfrac%7B4%2B2%20times%20%20%5Csqrt%7B-1%7D%20%20%7D%7B2%7D%3D%20%20%5Cfrac%7B6%20times%20%20%5Csqrt%7B-1%7D%20%20%7D%7B2%7D%3D3%20times%20%20%5Csqrt%7B-1%7D%20%5B%5Ctex%5D%5B%5Ctex%5D%5Csqrt%7B-1%7D%20%20)
representeds by 'i' so solution is 3i
then if other way around then wyou would do

=

and [\tex]\sqrt{-1} [/tex] is represented by i
the solution is x=3i or i (i=

)
but i is not real, it is imaginary so there are no real solution so the answer is C
Answer:
The answer is 70% if you round it.
Step-by-step explanation:
45 out of 68 is 66 but if you're rounding it the answer is 70%.
Answer:
(7-5i)( 2+3i)
=7(2)+ 7(3i)-5i(2) -(5i)(3i)
= 14 +21i -10i +15
= 29 +11i