All categories

3 years ago

How can I tell if the graph of an inequality will have a solid line?

Mathematics

1 answer:

love history [14]3 years ago

5 0

Answer:

When there's a line under < and >.

Step-by-step explanation:

You might be interested in

The graph of a proportional relationship passes through (12, 16) and (1, y) . Find y

k0ka [10]

Answer: The value of y is $\frac{4}{3}$ .

Explanation:

It is given that the graph of a proportional relationship passes through (12, 16)

and (1, y).

The graph of a proportional relationship means the x and y coordinates are in a proportion k. The equation of the graph is in the form of y=kx. Where k is the proportion factor.

It is given that the graph passing through (12,16).

$y=kx$

$16=k(12)$

$k=\frac{16}{12}$

$k =\frac{4}{3}$

So the equation of the line is,

$y=\frac{4}{3} x$

put x=1.

$y=\frac{4}{3} (1)$

$y=\frac{4}{3}$

Therefore, the value of y is $\frac{4}{3}$ .

8 0

3 years ago

Read 2 more answers

The roots of the quadratic equation $z^2 + az + b = 0$ are $2 - 3i$ and $2 + 3i$. What is $a+b$?

AlladinOne [14]

We know for our problem that the zeroes of our quadratic equation are $(2-3i)$ and $(2+3i)$ , which means that the solutions for our equation are $x=2-3i$ and $x=2+3i$ . We are going to use those solutions to express our quadratic equation in the form $a x^{2} +bx+c$ ; to do that we will use the zero factor property in reverse:
 $x=2-3i$
$x-2=-3i$
$x-2+3i=0$

 $x=2+3i$
$x-2=3i$
$x-2-3i=0$

Now, we can multiply the left sides of our equations:
 $(x-2+3i)(x-2-3i)= x^{2} -2x-3ix-2x+4+6i+3ix-6i-3^2i^2$
= $x^{2}-4x+4-9i^{2}$
= $x^{2} -4x+4+9$
= $x^{2} -4x+13$
Now that we have our quadratic in the form $a x^{2} +bx+c$ , we can infer that $a=1$ and $b=-4$ ; therefore, we can conclude that $a+b=1+(-4)=1-4=-3$ .