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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Area = Length x Width
63 = 9 x width
Width= 63 ÷ 9 = 7 feet
Answer: 7 feet
Answer:
y = 3/5 x + 2.
Step-by-step explanation:
Use the point-slope equation of a straight line:
y - y1 = m (x - x1) where m = the slope amd (x1, y1) is a point on the line.
Here:
m = (5- (-1)) / (5 - (-5))
= 6/10
= 3/5.
Substituting for m and (5, 5):
y - 5 = 3/5(x - 5)
y - 5 = 3/5x - 3
y = 3/5 x + 2.
Answer:
Area: ½ × base × height
Perimeter: sum of side lengths of the triangle
of vertices: 3
Number of edges: 3
Internal angle: 60° (for equilateral)
Sum of interior angles: 180°
Step-by-step explanation: