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4 years ago

11

Which statement best illustrates using the vertical line test to determine if the graph below is a function of x ?

2 answers:

kirill115 [55]4 years ago

4 0

If something is a function, every x-value, only has one (1) y-value. If the line x = 0 intersects the graph in two points, there are to values for y (instead of one). Therefore the graph is not a function of x.

Your answer is A.

Mekhanik [1.2K]4 years ago

3 0

Answer:

A

Step-by-step explanation:

A function can only have one output value "y" for each input "x", so if a vertical line in an xy plane (therefor the line test) that you can imagine, or you can see by using a ruler intersects more than one point on the same line, you have more than one value for x, and that is not a function.

as you see in option A, the line x=0 intersects the graph at two points.

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2. What are the values of a and b?<br> (1 point)

IRINA_888 [86]

Answer:

9 13

a_. b=_

3 3

Step-by-step explanation:

it's that because you can devid 9 by 3

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3 years ago

Answer quickly please!

LenaWriter [7]

1. D
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4 years ago

26. Define a relation ∼ ∼ on R 2 R2 by stating that ( a , b ) ∼ ( c , d ) (a,b)∼(c,d) if and only if a 2 + b 2 ≤ c 2 + d 2 . a2+

Tresset [83]

Answer:

~ is reflexive.

~ is asymmetric.

~ is transitive.

Step-by-step explanation:

~ is reflexive:

i.e., to prove $\forall (a, b) \in \mathbb{R}^2$ , $(a, b) R(a, b)$ .

That is, every element in the domain is related to itself.

The given relation is $\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$

Reflexive:

$(a, b) \sim (a, b)$ since $a^2 + b^2 = a^2 + b^2$

This is true for any pair of numbers in $\mathbb{R}^2$ . So, $\sim$ is reflexive.

Symmetry:

$\sim$ is symmetry iff whenever $(a, b) \sim (c, d)$ then $(c, d) \sim (a, b)$ .

Consider the following counter - example.

Let (a, b) = (2, 3) and (c, d) = (6, 3)

$a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13$

$c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42$

Hence, $(a, b) \sim (c, d)$ since $a^2 + b^2 \leq c^2 + d^2$

Note that $c^2 + d^2 \nleq a^2 + b^2$

Hence, the given relation is not symmetric.

Transitive:

$\sim$ is transitive iff whenever $(a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f)$ then $(a, b) \sim (e, f)$

To prove transitivity let us assume $(a, b) \sim (c, d)$ and $(c, d) \sim (e, f)$ .

We have to show $(a, b) \sim (e, f)$

Since $(a, b) \sim (c, d)$ we have: $a^2 + b^2 \leq c^2 + d^2$

Since $(c, d) \sim (e, f)$ we have: $c^2 + d^2 \leq e^2 + f^2$

Combining both the inequalities we get:

$a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2$

Therefore, we get: $a^2 + b^2 \leq e^2 + f^2$

Therefore, $\sim$ is transitive.

Hence, proved.

3 0

3 years ago

which term describes the point where the perpendicular bisectors of the three sides of a triangle intersect?

aleksklad [387]

Circumcenter for the Apex students :)

6 0

3 years ago

Please help me 123 or 4

butalik [34]

Answer:

3

Step-by-step explanation:

because divide Pt from both sides to get r=i/(pt)

8 0

3 years ago