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3 years ago

Which relations are functions? Select Function or Not a function for each graph.

Mathematics

1 answer:

Alisiya [41]3 years ago

3 0

Answer:

A and C are not functions, B and D are functions.

Step-by-step explanation:

If you were to stick a vertical line on the graph, and it passes through the line of the graph more than once, it is not a function.

I hope this helps :)

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HELP! will mark BRAINILEST !

Sedbober [7]

I like A as the best one

420 = 100*price - (fixed costs)
220 = 60*price - (fixed costs)

If we subtract one from the other one, we get 200 = 40*price, price = 5
that said, fixed costs are 500 - 420 = 80

Now if we move -220 to the right in Answer A, we'll get:

y = 5x -300 + 220 = 5x - 80

So it looks like A describes the model best

6 0

3 years ago

Richard claims that the value of the expression 23x2+173x+12 is sometimes negative. Which of these values of x supports Richard'

Talja [164]

The correct answer is B.

6 0

2 years ago

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What is the quotient of (a2 +6a + 9) +(a + 3)

bogdanovich [222]

Im sorry but this is not a quotient problem but an addition one, I can still simplyfy it for you though.

answer : $a^2+7a+12$

6 0

3 years ago

If x+a is a factor of 2x²+2ax+5x+10,find a.

vazorg [7]

If $x+a$ is a factor of the polynomial, it means that $-a$ is a root of the polynomial.

In fact, if a polynomial $p(x)$ has a root $x_0$ (i.e. $p(x_0)=0$ ), then the polynomial is divisible by $(x-x_0)$

In your case, you know that the polynomial is divisible by $(x+a)=(x-(-a))$ , so $-a$ is a solution.

Let's evaluate $p(-a)$ and set it to zero:

$2(-a)^2+2a(-a)+5(-a)+10 = 0 \iff 2a^2-2a^2-5a+10=0 \iff -5a+10 = 0 \iff a = 2$

We can check the answer: if $a=2$ the polynomial becomes

$2x^2+2(2)x+5x+10 = 2x^2+9x+10$

And we're claiming that $-a=-2$ is a root of this polynomial:

$2(-2)^2+9(-2)+10 = 2\cdot 4 - 18 + 10 = 0$

5 0

3 years ago

Use the graph below to fill in the blank with the correct number:

Alik [6]

Let
$A(-3,0) \\ B(-2,2) \\ C(0,1) \\ D(1,-2)$

we know that

Point A
$A(-3,0) \\ for x=-3 \\ f(-3)=0$

Point B
$B(-2,2) \\ for x=-2 \\ f(-2)=2$

Point C
$C(0,1) \\ for x=0 \\ f(0)=1$

Point D
$D(1,-2) \\ for x=1 \\ f(1)=-2$

therefore

the answer is
$f(0)=1$

4 0

3 years ago

Read 2 more answers