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:
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Answer:
Step-by-step explanation:
Integer roots:
a = 6 n + 1, b = 7, c = -3 n^2 - n, n element Z
a = 6 n + 5, b = 7, c = -3 n^2 - 5 n - 2, n element Z
Derivative:
d/da(1/7 (a^2 + 12 c) b - 1) = (2 a b)/7
Indefinite integral:
integral(-1 + 1/7 b (a^2 + 12 c)) da = (a^3 b)/21 + (12 a b c)/7 - a + constant
Related Queries:
Propportinal
means that the points are inproportion
ratios
1:3=2:6=3:7
1/3=2/6=3/7
1/3=1/3=3/7
false
therefor it is not poropoortional
another way is to graph the points and see if a straight line can go though all of them
it can go thoruhg the first 3 but not eh 4th pont (3,7)
not proportional
The measure of an arc refers to the arc length divided by the radius of the circle. The arc measure equals the corresponding central angle measure, in radians. That's why radians are natural: a central angle of one radian will span an arc exactly one radius long.
Is the correct
Answer: (7, 20)
Concept:
There are three general ways to solve systems of equations:
- Elimination
- Substitution
- Graphing
Since the question has specific requirements, we are going to use <u>substitution </u>to solve the equations.
Solve:
<u>Given equations</u>
y = 3x - 1
2x + 6 = y
<u>Substitute the y value since both equations has isolated [y]</u>
2x + 6 = 3x - 1
<u>Add 1 on both sides</u>
2x + 6 + 1 = 3x - 1 + 1
2x + 7 = 3x
<u>Subtract 2x on both sides</u>
2x + 7 - 2x = 3x - 2x
<u>Find the value of y</u>
y = 3x - 1
y = 3(7) - 1
y = 21 - 1
Hope this helps!! :)
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