Does anyone know how to do these? I'm having a hard time and you guys are very helpful.

Mathematics

2 answers:

Keith_Richards [23]2 years ago

6 0

4z - 1 ( for now skip the z part) = 3z.

3z times 2 = 6z

Now on the other side, add the 3 outside of the parentheses to the 2 on the inside... you'll get 5 + z

To make each side equal... 5 + ? = 6

Obviously 1... So z = 1

n200080 [17]2 years ago

4 0

Ok so keep in mind, we are solving for x or z. We want to get it by itself.

So, with 6x= 5x+22, we need to subtract 5x from both sides of the = sign

So 6x - 5x = 1x.

Our new equation looks like 1x= 22

We still don't have x alone so we need to divide one on both sides of the =

Our new and final answer is x=22

You might be interested in

Rationalize the denominator of $\frac{5}{2+\sqrt{6}}$. The answer can be written as $\frac{A\sqrt{B}+C}{D}$, where $A$, $B$, $C$

horrorfan [7]

Answer:

$A +B+C+D = 3$ is the correct answer.

Step-by-step explanation:

Given:

$\frac{5}{2+\sqrt{6}}$

To find:

$A+B+C+D = ?$ if given term is written as following:

$\frac{A\sqrt{B}+C}{D}$

Solution:

We can see that the resulting expression does not contain anything under $\sqrt$ (square root) so we need to rationalize the denominator to remove the square root from denominator.

The rule to rationalize is:

Any term having square root term in the denominator, multiply and divide with the expression by changing the sign of square root term of the denominator.

Applying this rule to rationalize the given expression:

$\dfrac{5}{2+\sqrt{6}} \times \dfrac{2-\sqrt6}{2-\sqrt6}\\\Rightarrow \dfrac{5 \times (2-\sqrt6)}{(2+\sqrt{6}) \times (2-\sqrt6)} \\\Rightarrow \dfrac{10-5\sqrt6}{2^2-(\sqrt6)^2}\ \ \ \ \ (\because \bold{(a+b)(a-b)=a^2-b^2})\\\Rightarrow \dfrac{10-5\sqrt6}{4-6}\\\Rightarrow \dfrac{10-5\sqrt6}{-2}\\\Rightarrow \dfrac{-5\sqrt6+10}{-2}\\\Rightarrow \dfrac{5\sqrt6-10}{2}$