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3 years ago

HELP ITS ALGEBRA 2!!

Mathematics

1 answer:

Orlov [11]3 years ago

5 0

To find the rate of change, do:
$\frac{ y_{1} - y_{2} }{ x_{1} - x_{2} }$

1) 4-2 / 2-1 = 2

2) 8-4 / 3-2 = 4

3) 16-8 / 4-3 = 8

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Is this a linear function , why or why not ? explain <br> y=x^2-1

kotykmax [81]

If you graph out y = x^2 - 1, you'll see that it is a curved parabola. It is not a straight line. So visually this shows the equation is not linear. It's considered nonlinear.

Algebraically, the given equation does not fit the form y = mx+b which is slope intercept form. So this is more confirmation that we do not have a linear function.

8 0

3 years ago

What are the coordinates of point B on \overline{AC} such that AB = 1/2 BC

marta [7]

Answer:

C. $B(x,y) = \left(3,-\frac{16}{3} \right)$

Step-by-step explanation:

Let $A(x,y) = (6,-7)$ , $B(x,y) = (x,y)$ and $C(x,y) = (-3,-2)$ . From statement we know that $AB = \frac{1}{2}\cdot BC$ , which is equivalent to the following linear algebraic formula:

$B(x,y) -A(x,y) = \frac{1}{2}\cdot [C(x,y)-B(x,y)]$ (1)

$B(x,y)-A(x,y) = \frac{1}{2}\cdot C(x,y)-\frac{1}{2}\cdot B(x,y)$

$\frac{3}{2}\cdot B(x,y) = \frac{1}{2}\cdot C(x,y)+A(x,y)$

$B(x,y) = \frac{1}{3}\cdot C(x,y) +\frac{2}{3}\cdot A(x,y)$ (2)

Then, the coordinates of point B on AC are:

$B(x,y) = \frac{1}{3}\cdot (-3,-2)+\frac{2}{3}\cdot (6,-7)$

$B(x,y) = \left(-1, -\frac{2}{3}\right)+\left(4, -\frac{14}{3} \right)$

$B(x,y) = \left(3,-\frac{16}{3} \right)$

Which means that correct answer is C.

5 0

3 years ago

I will give brainliest if correct

dalvyx [7]

Answer:supplementary

Step-by-step explanation:

Opposite angles in any quadrilateral inscribed in a circle are supplements of each other.

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3 years ago

The graph of which function will have a maximum and a y-intercept of 4? f(x) = 4x2 + 6x – 1 f(x) = –4x2 + 8x + 5 f(x) = –x2 + 2x

V125BC [204]

Answer:

Option C (f(x) = $-x^2 + 2x + 4$ )

Step-by-step explanation:

In this question, the first step is to write the general form of the quadratic equation, which is f(x) = $ax^2 + bx + c$ , where a, b, and c are the arbitrary constants. There are certain characteristics of the values of a, b, and c which determine the nature of the function. If a is a positive coefficient (i.e. if a>0), then the quadratic function is a minimizing function. On the other hand, a is negative (i.e. if a<0), then the quadratic function is a maximizing function. Since the latter condition is required, therefore, the first option (f(x) = $4x^2 + 6x - 1$ ) and the last option (f(x) = $x^2 + 4x - 4$ ) are incorrect. The features of the values of b are irrelevant in this question, so that will not be discussed here. The value of c is actually the y-intercept of the quadratic equation. Since the y-intercept is 4, the correct choice for this question will be Option C (f(x) = $-x^2 + 2x + 4$ ). In short, Option C fulfills both the criteria of the function which has a maximum and a y-intercept of 4!!!