Which points are separated by distance of 6 units?

Mathematics

1 answer:

sergiy2304 [10]3 years ago

8 0

I think it’s b! i hope that helps!

Read more about graphs at:

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7 0

2 years ago

Anyone knowww? :) help!

Yuki888 [10]

The answer would be 17.
This is because, the format for finding out the hypotenuse of a triangle is--
a^2 + b^2 = c^2
So, it would translate to--
15^2 + 8^2 = c^2
225 + 64 = 289
Because 289 = c^2, you need to take the square root of it.
The square root of 289 is 17.
Therefore, your answer is 17.

8 0

3 years ago

Read 2 more answers

Which solution to the equation 1/x-1=x-2/2x^2-2 is extraneous?

olga2289 [7]

An extraneous solution is a solution, such as that to an equation, that emerges from the process of solving the problem but is not a valid solution to the original problem.

Given the equation:
$\frac{1}{x-1} = \frac{x-2}{2x^2-2} \\ \\ 2x^2-2=(x-1)(x-2)=x^2-3x+2 \\ \\ x^2+3x-4=0 \\ \\ (x-1)(x+4)=0 \\ \\ x-1=0 \ or \ x+4=0 \\ \\ x=1 \ or \ x=-4$

From the equation, it is obtained that x = 1 and x = -4 are the solutions of the equation.

Now, we substitute x = 1 into the equation as follows:
 $\frac{1}{1-1} = \frac{1-2}{2(1)^2-2} \\ \\ \frac{1}{0} = \frac{-1}{2-2} \\ \\ \frac{1}{0} = \frac{-1}{0}$

As can be see, for x = 1, the equation is undefined.

Now, we substitute x = -4 into the equation as follows:
$\frac{1}{-4-1} = \frac{-4-2}{2(-4)^2-2} \\ \\ \frac{1}{-5} = \frac{-6}{2(16)-2} = \frac{-6}{32-2} = \frac{-6}{30} \\ \\ - \frac{1}{5} =- \frac{1}{5}$

It can be seen that x = -4 is a valid solution of the orignal equation.

Therefore, x = 1 is an extraneous solution to the equation
$\frac{1}{x-1} = \frac{x-2}{2x^2-2}$

8 0

3 years ago

Solve (x+1<4)∩(x-8>-7)

tino4ka555 [31]

We can rewrite the two sets as

$x+11$