<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
:)</span>
<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:
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.
Answer:
The and is x-2)-2(x + 3)3=0