What is the range of function of y= 9x
Just evaluate the function where the value passed to the delta is 0. i.e. if your are trying to integrate x^2*delta(x-3)dx, that is just equal to the value of x^2=3^2=9 since x-3=0 at x=3. If the limits of integration do not include the value where delta is 0, then the integral is 0 since delta(x)=0 everywhere that x is not=0. Thinking of it from a graphical perspective, you are asking for the area under the curve of a function multiplied by the delta function, which just leaves the portion of the graph at where the spike from delta happens. Everywhere else, the graph is 0. So the only thing that contributes to the integral is the value of the function where delta(0) happens. Since the integral of the function at that point is constant and delta at that point is just 1, it's just the value of the function at that point. I do not believe there is a delta function in the TI-NSpire for you to do this directly. You need to recognized the meaning of the delta function.
2x2 - 12x + 18 = 0 (divide by 2 both sides) x2 - 6x + 9 = 0 (the double root is 3) (x - 3)2 = 0 x - 3 = 0 (add 3 to both sides) x = 3 Thus, the zero of f(x) = 2x2 - 12x + 18 is 3.
three 2 - 2 = 0 0 x 3 = 0 0 + 3 = 3
y = 1/2 x + 1
5.71
93 = 729
yes
Y = x2
If those are the only values, no.
The domain is the set {-3, -2, 0, 3}. Note that because -2 is mapped to -5 as well as 6, this relationship is not a function.
(2, 4)
The Equation of a Rational Function has the Form,... f(x) = g(x)/h(x) where h(x) is not equal to zero. We will use a given Rational Function as an Example to graph showing the Vertical and Horizontal Asymptotes, and also the Hole in the Graph of that Function, if they exist. Let the Rational Function be,... f(x) = (x-2)/(x² - 5x + 6). f(x) = (x-2)/[(x-2)(x-3)]. Now if the Denominator (x-2)(x-3) = 0, then the Rational function will be Undefined, that is, the case of Division by Zero (0). So, in the Rational Function f(x) = (x-2)/[(x-2)(x-3)], we see that at x=2 or x=3, the Denominator is equal to Zero (0). But at x=3, we notice that the Numerator is equal to ( 1 ), that is, f(3) = 1/0, hence a Vertical Asymptote at x = 3. But at x=2, we have f(2) = 0/0, 'meaningless'. There is a Hole in the Graph at x = 2.
No, it is not a function.
10% of 30 is 3 9 = 3 x 3 So 3 x 10% = answer 30%
If: x = -2 and x = 3/4 Then: (4x-3)(x+2) = 0 So: 4x2+5x-6 = 0
2 sin(x) - 3 = 0 2 sin(x) = 3 sin(x) = 1.5 No solution. The maximum value of the sine function is 1.0 .