Answer:
5 but its probaly not right bc im just looking for pointsStep-by-step explanation:
Answer:
D) (0, -3)
Step-by-step explanation:
The solution of two linear equations is the point of intersection of their graphs.
From the graph, the point of intersection of the graphs of the two linear equations is on the y-axis. So, the value on the axis is always 0. Also, the point is below the x-axis. So, the value of the point will be negative.
From the choices given, only (0,-3) matches the above conditions.
Therefore, the correct choice is D. (0, -3).
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
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The Minimum sample size table is attached below
Answer:
Step-by-step explanation:
From the question we are told that:
Confidence Interval
Variance
Generally going through the table the
Minimum sample size is
Given the coordinates of the image of line segment RT to be R'(-2,-4) and T'(4.4), if the image produced was dilated by a scale factor of 12 centered at the origin, to get the coordinate of the end point, we will simply multiply the x and y coordinates of by the factor of 12 as shown:
For R' with coordinate R'(-2,-4), the coordinates of endpoint of the pre-image will be:
R = 12R'
R = 12(-2, -4)
R = (-24, -48)
For T' with coordinate T'(4,4), the coordinates of endpoint of the pre-imagee will be:
T = 12T'
T = 12(4, 4)
T = (48, 48)
Hence the coordinate of the endpoint of the preimage will be at R(-24, -48) and T(48, 48)