Helppppp, i dont know how to do this problem

Mathematics

1 answer:

Nana76 [90]3 years ago

5 0

Answer:

(3, -5) (4, -5) (2, -7)

Step-by-step explanation:

An integer is any whole number that is not a fraction, they include -10204, -3, 0, 5, and 8592. (random numbers)

A coordinate with integers can be anything such as (0, 4582) and (547, 42)

However, it can not be (0.45846, 4579) or (54.42, 58).

To determine which points have coordinates that look like they are integers, you just have to select 3 points that lie at the intersection of the y and x value tic marks

You can see that (3, -5) is at the intersection of the x value 3 tic mark and the y value -5 tic mark, while also being on the parabola.

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In simplest radical form, what are the solutions to the quadratic equation 6 = x2 – 10x?

vichka [17]

Answer:

$x = 5+\sqrt{31}\,\, and\,\, x=5-\sqrt{31}$

Step-by-step explanation:

We need to solve the quadratic equation

6 = x^2 -10x

Rearranging we get,

x^2-10x-6=0

Using quadratic formula to solve the quadratic equation

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b =-10 and c=6

Putting values in the quadratic formula

$x=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(-6)}}{2(1)}\\x=\frac{10\pm\sqrt{100+24}}{2}\\x=\frac{10\pm\sqrt{124}}{2}\\x=\frac{10\pm\sqrt{2*2*31}}{2}\\x=\frac{10\pm\sqrt{2^2*31}}{2}\\x=\frac{10\pm2\sqrt{31}}{2}\\x = 5\pm\sqrt{31}$