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.
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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.
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Answer: 15(5z^2+4z+8)
Step-by-step explanation:
You take out the common values
Answer:
Your brother can buy anywhere from 1 to 6 baseball cards depending on how rare they are!
Vertex = (3, - 2) this should be your answer. dm if im wrong
The negative sign shows that they still owed $5000 to the creditors which shows a loss of money between June and December.
Step-by-step explanation:
Amount owed at the end of June = $3765
Amount owed at the end of December = $8765
As Hightop Roofing owed creditors the amount, therefore, we will consider the amount as negative, therefore,
Amount in June = -3765
Amount in December = -8765
To see the difference, we will subtract the amount owed in June from the amount owed in December.
The negative sign shows that they still owed $5000 to the creditors which shows a loss of money between June and December.
Keywords: difference, loss
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