M can be any positive real number.
Explanation:
From f(x) = √(mx) ; if x is posive m has to be positive; if x is negative m has to be negative; if x is cero m can have any value, and the range will always be 0 or positve
From g(x) = m √x; x can only be 0 or positive and the range will have the sign of m.
Given we concluded that the range of f(x) can only be 0 or positive, then me can only be 0 or positive.
42 can be
1 x 42, or
2 x 21, or
3 x 14, or
6 x 7 .
Answer:
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Find a point-slope form for the line that satisfies the stated conditions. Slope , passing through (-5,4)
I really need this question answered
By:
I don't see a value for the slope. We need that to set the equation, otherwise I can write an unlimited number of equations that pass through (-5,4).
I'll assume a slope so that you can see how the procedure would work. I like 6, so we'll assume a slope of 6.
The equation for a straight line has the form y = mx + b, where m is the slope and y is the y-intercept, the value of y when x = 0. We want a line that has slope 6, so:
y = 6x + b
We need to find b, so substitute the point (-5,4) that we know is on the line:
4 = 6*(-5) + b and solve for b
4 = -30 + b
b = 34
The line is y = 6x + 34
