Answer:
D and F
Step-by-step explanation:
Please mark brainliest.
You would cross multiply 53 times 5 and divide by 1
<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
1/2 times 6 = 1/2 * 6
so in order to make this multiplication you have to turn 6 to a fraction, 6 is equal to 6/1
so , the way to multiply them is multiply the top with the top and it will be the top in the answer and the exact same with the bottoms
6*1= 6
2*1=2
so the fraction is 6/2, since a fraction is basicly a division
6/2 = 3
so the answer to your question is 3
i hope i helped you
Note that the 2nd equation can be re-written as y=8x-10.
According to the second equation, y=x^2+12x+30.
Equate these two equations to eliminate y:
8x-10 = x^2+12x+30
Group all terms together on the right side. To do this, add -8x+10 to both sides. Then 0 = x^2 +4x +40. You must now solve this quadratic equation for x, if possible. I found that this equation has NO REAL SOLUTIONS, so we must conclude that the given system of equations has NO REAL SOLUTIONS.
If you have a graphing calculator, please graph 8x-10 and x^2+12x+30 on the same screen. You will see two separate graphs that do NOT intersect. This is another way in which to see / conclude that there is NO REAL SOLUTION to this system of equations.