All categories

3 years ago

If f is a function and x is an element in its domain, which statement is true about the graph of f?

1 answer:

8 0

An appropriate choice among these is
The graph of f is the graph of the equation y = f(x).

_____
Horizontal and vertical axes of the Cartesian plane are conventionally labeled and referred to as "x" and "y", respectively. When we talk about the graph of f(x) in that context, we usually mean the graph of y=f(x). However, this convention may not be followed in all cases. There may be no "y" label on the graph at all, or the horizontal axis may be labeled something other than "x".

You might be interested in

PLEASE HELP!! What are the explicit equation and domain for a geometric sequence with the first term of 3 a second term of -9?

frutty [35]

Answer:

The correct answer is c.

4 0

3 years ago

Read 2 more answers

What is this this sos 3x-y=3? (x,y)=?

Mazyrski [523]

Rewrite the equation;

3x-y=3
3x-3=y

Linear equation.

I'm unsure of what the second part asks. Can you rewrite the given question?

6 0

3 years ago

If for two non-zero vectors a,b : ∣a+b∣=∣a−b∣ then angle between them is:______a. 0b. 45c. 60d. 90

Eddi Din [679]

Answer:

$90^\circ$

Step-by-step explanation:

Given two non zero vectors, $\vec a, \vec b$ .

Let the angle between the two vectors = $\theta$

Given that:

$|\vec a+\vec b|=|\vec a-\vec b|$

Let us have a look at the formula for magnitude of addition of two vectors:

$|\vec x+\vec y|=\sqrt{x^2+y^2+2xycos\theta}$

Where $\theta$ is the angle between the two vectors.

formula for magnitude of subtraction of two vectors:

$|\vec x-\vec y|=\sqrt{x^2+y^2-2xycos\theta}$

As per the given condition:

$\sqrt{a^2+b^2+2abcos\theta}=\sqrt{a^2+b^2-2abcos\theta}$

Squaring both sides:

$a^2+b^2+2abcos\theta=a^2+b^2-2abcos\theta\\\Rightarrow 2abcos\theta=-2abcos\theta\\\Rightarrow cos\theta = -cos\theta\\\Rightarrow 2cos\theta = 0\\\Rightarrow cos\theta = 0\\\Rightarrow \theta = 90^\circ$

So, the angle between the two vectors is: $90^\circ$

7 0

3 years ago

Please help! a rectangle's width is 6 feet less than it's length. write a quadratic function that expresses the rectangles area

Monica [59]

Answer:

B. A( L) = L² - 6L

Step-by-step explanation:

A rectangle has 2 sides, length and width.

length = L

width = L - 6

Area = length × width

= L(L - 6)

= L² - 6L

The correct choice is B.

B. A( L) = L² - 6L

7 0

3 years ago

Read 2 more answers

Let A = (1, 0, 3) and B = (−3, 2, 1).

Romashka-Z-Leto [24]

Answer:

a) $(-\frac{\sqrt{24}}{6} ,\frac{\sqrt{24}}{12} ,-\frac{\sqrt{24}}{12})$

b) $(-1,1,2)$

c) $-2x+y-z=1$

Step-by-step explanation:

AB, OB, OA are vectors.

$AB = OB - OA = (-3, 2, 1) - (1, 0, 3) = (-4, 2, -2)\\||AB|| = \sqrt{(-4)^2+2^2+(-2)^2}=\sqrt{24}\\$

Normalizing AB: $\frac{AB}{||AB||} = (\frac{-4}{\sqrt{24}},\frac{2}{\sqrt{24}}, \frac{-2}{\sqrt{24}})=(-\frac{\sqrt{24}}{6} ,\frac{\sqrt{24}}{12} ,-\frac{\sqrt{24}}{12})$

OM, OA, OB are vectors.

$OM = \frac{1}{2}OA+\frac{1}{2}OB=(\frac{-3+1}{2} ,\frac{2+0}{2}, \frac{1+3}{2} )=(-1,1,2)$

So, $M = (-1,1,2)$

$-4x+2y-2z=(-4)(-1)+(2)(1)+(-2)(2)=4+2-4=2$

Hence,

$-2x+y-z=1$

5 0

3 years ago