Build the graph of the equation (x+8)(y-10)=0 FIND THE TWO POINTS

Mathematics

1 answer:

alisha [4.7K]3 years ago

3 0

Answer:

Step-by-step explanation:

Theoretically speaking, both equations need to equal 0.

Let's start with (x + 8)

x + 8 = 0

x = 0 - 8

x = -8

Then, let's go to (y - 10)

y - 10 = 0

y = 10 = 0

y = 10

So, x = -8 & y = 10

Hope that helps!

You might be interested in

If Vector u=(5,-7) and v=(-11,3), 2v-6u=_____ and ||2v-6u||≈_____

victus00 [196]

<u>Part 1</u>: Correct.

<u /> $\sqrt{(-52)^2 + 48^2} \approx 70.77$

8 0

2 years ago

Whats the difference between graphing and writing equations from table, points, and equations,

satela [25.4K]

The difference is that a graph is multiple of them and a table is in sections in points it’s like jotting them down and equations is step-by-step

Hope the help:)

5 0

3 years ago

Which equation could generate the curve in the graph below?y=3x^2-2x+1

4vir4ik [10]

If you are asking what is the graph of y = 3x^2 -2x+1.

Then, the attached file would be the answer.

To check, b^2 - 4(a)(c), for each equation and use these facts:

If b^2 - 4(a)(c) = 0, there is only one real root meaning, the graph touches the x-axis only in one point.

If b^2 - 4ac > 0, there are two real roots meaning, the graph touches the x-axis in two different points.

If b2 - 4ac < 0, there are no real roots then the graph does not touch the x-axis. This would be the case for y = 3x^2 - 2x + 1.
Solution:
(-2)^2 -4(3)(1) = 4 - 12 = -8 < 0 will result in not real roots.