What is the first step when solving this equation?

Mathematics

2 answers:

notka56 [123]3 years ago

6 0

Answer:

Distribute

When you have an equation like this, distribute first and then add or subtract like terms to solve for x.

Hope this helps!!

snow_tiger [21]3 years ago

3 0

Well, you need to find out what x is equal to, if anything.

The first step to completing it, from what I've learned, is to do the problem that is in parenthesis first.

Hope this helps, sorry if not-

You might be interested in

I want to know the answer to the sum of f and 6

11111nata11111 [884]

f + 6 = ?

Well, the answer can be pretty much any number. It all depends on what number f is. f is a variable, so you can plug in any number to replace f.

For example, if f was 5, this is what the equation would look like:

5 + 6 = 11

If f was -3, this is what the equation would look like:

-3 + 6 = 3

So, again, the answer can be anything. It depends on what f is.

4 0

3 years ago

Please show me step by step how to solve this problem 3(-8+4v)=8v ?

Nadusha1986 [10]

So we know that:
3(-8 + 4v) = 8v
To find v, first simplify the left hand side:
-24 + 12v = 8v
Then group the "v's" over to the right:
-24 = -4v
-6 = -v
So v = 6

Hope this helped

8 0

3 years ago

The points A BC :(2, 2,1), :(1,1,3), :(2,0,5) − are the vertices of a right triangle. The radius of the sphere with center at th

ElenaW [278]

Answer:

r=1

Step-by-step explanation:

First we need to know the length of each side of the triangle, so we use the formula of the vector modulus:

$|AB|= \sqrt{(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}+(b_{3}-a_{3})^{2}$

By doing so, we find:

$|AB|=\sqrt {6}\\|BC|=\sqrt {6}\\|AC|=2\sqrt {5}$

With this we know that the triangle is not right, but, we assume the longest side as the hypotenuse of the problem.

As we have two equal sides, we know that the line between point |AB| and the center of the hypotenuse is perpendicular, therefore, we can calculate it using Pythagoras theorem:

$|BC|^{2}=r^{2}+(\frac{|AC|}{2})^{2}\\\\r^{2}=|BC|^{2}-(\frac{|AC|}{2})^{2}\\\\r^{2}=(\sqrt{6})^{2}- (\frac{2\sqrt{5}}{2})^{2}\\\\r^{2}=6-5=1\\r=1$