Answer: jeygehrhrxftjh
Step-by-step explanation:
Answer:
see below
Step-by-step explanation:
The function f(x) is the absolute value function scaled vertically and shifted up. Neither of those transformations affects the interval on which it is increasing. The absolute value function increases for x > 0. (It is neither increasing nor decreasing at x=0.)
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The function g(x) is a parabola that opens downward. Consequently, it is increasing for all values of x to the left of its vertex (or line of symmetry). That line of symmetry can be found as ...
x = -b/(2a) = -(8)/(2·(-1)) = 4
So g(x) is increasing from -∞ to 4, but at x=4 is neither increasing nor decreasing.
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Then the interval on which both functions are increasing is ...
{x > 0} ∩ {x < 4} = {0 < x < 4}
This will be graphed as a line between 0 and 4 with open circles at each end.
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The second attachment shows the two functions so you can see where their slopes are positive.
The question seems to be pointing towards a multiple choice response but I will do my best. The first equation would be 15x+12y=9.30 and the second would be 10x+4y=4.60, x represents flour and y represents sugar.
Hope this helps.
Answer:
Step-by-step explain
Find the horizontal asymptote for f(x)=(3x^2-1)/(2x-1) :
A rational function will have a horizontal asymptote of y=0 if the degree of the numerator is less than the degree of the denominator. It will have a horizontal asymptote of y=a_n/b_n if the degree of the numerator is the same as the degree of the denominator (where a_n,b_n are the leading coefficients of the numerator and denominator respectively when both are in standard form.)
If a rational function has a numerator of greater degree than the denominator, there will be no horizontal asymptote. However, if the degrees are 1 apart, there will be an oblique (slant) asymptote.
For the given function, there is no horizontal asymptote.
We can find the slant asymptote by using long division:
(3x^2-1)/(2x-1)=(2x-1)(3/2x+3/4-(1/4)/(2x-1))
The slant asymptote is y=3/2x+3/4