5 Simple Algerba 2 Questions.

Mathematics

1 answer:

zubka84 [21]3 years ago

6 0

Answer:

See below for answers

Step-by-step explanation:

1) Parent functions are essentially non-transformed functions, so y=x^2 would be the parent function of y=x^2+1 for example since y=x^2 has no transformations while y=x^2+1 does.

2) Since certain x-values won't be defined for the function, you need to restrict the domain. For example, in the function f(x)=1/x, obviously x=0 is undefined for the function, so we would need to restrict the domain.

3) Not sure what you mean by name graph; no context given

4) Also no context given

5) The equation would be y=(x+5)^4-3

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Find the length of the following two-dimensional curve. r (t ) = (1/2 t^2, 1/3(2t+1)^3/2) for 0 < t < 16

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Answer:

r = 144 units

Step-by-step explanation:

The given curve corresponds to a parametric function in which the Cartesian coordinates are written in terms of a parameter "t". In that sense, any change in x can also change in y owing to this direct relationship with "t". To find the length of the curve is useful the following expression;

$r(t)=\int\limits^a_b ({r`)^2 \, dt =\int\limits^b_a \sqrt{((\frac{dx}{dt} )^2 +\frac{dy}{dt} )^2)} dt$

In agreement with the given data from the exercise, the length of the curve is found in between two points, namely 0 < t < 16. In that case a=0 and b=16. The concept of the integral involves the sum of different areas at between the interval points, although this technique is powerful, it would be more convenient to use the integral notation written above.

Substituting the terms of the equation and the derivative of r´, as follows,

$r(t)= \int\limits^b_a \sqrt{((\frac{d((1/2)t^2)}{dt} )^2 +\frac{d((1/3)(2t+1)^{3/2})}{dt} )^2)} dt$