The weight in 1980 is
kilograms
<em><u>Solution:</u></em>
From 1980 to 1990, Lior’s weight increased by 25%
His weight is "k" kilograms in 1990
<em><u>To find: weight in 1980</u></em>
This is a percentage increase problem
Let "x" be the weight in kilograms in 1980
<em><u>The percentage increase is given by formula:</u></em>

Here,
Initial value in 1980 = x
Final value in 1990 = k
Percentage increase = 25 %
<em><u>Substituting the values in formula,</u></em>

Thus the weight in kilograms in 1980 is 
You can use a diffrent kind of ruler for math then you could find the answer.
Answer:
So, the step1 is correct.
Step-by-step explanation:
The expression is

So, the step 1 is correct.
By using the rules that the value inside square root can’t be negative and the denominator value can’t be zero, the domain for the given function is a) x<-1 and x>1 b) p≤1/2 c) s>-1.
I found the complete question on Chegg, here is the full question:
Write the restrictions that should be imposed on the variable for each of the following function. Then find, explicitly, the domain for each function and write it in the interval notation a) f(x)=(x-2)/(x-1) b) g(p)=√(1-2p) c) m(s)= (s^2+4s+4)/√(s+1)
Ans. We know that a number is not divisible by zero and number inside a square root can not be negative. In both the cases the outcome will be imaginary.
a) For this case the denominator x-1 can not be zero. So, x ≠1 and the domain is x<-1 and x>1.
b) For this case the value inside square root can’t be negative. So, p can’t be greater than 1/2 the domain is p≤1/2.
c) For this case also the value inside square root can’t be negative and the denominator value can’t be zero. So, s can’t equal or less than -1 and domain is s>-1.
Learn more about square root here:
brainly.com/question/3120622
#SPJ4
Answer:
H(s)=(∫_(t=o)^∞▒〖x(t)e^(-st) dt〗)/(∫_(t=o)^∞▒〖y(t) e^(-st) dt〗)
Step-by-step explanation:
L{f(t)}=F(s)=∫_(t=0)^∞▒〖f(t)e^(-st) dt〗