All categories

2 years ago

3x-3=2x Please help me...? 10 points given

2 answers:

6 0

3x - 3 = 2x

Add 3 on both sides

3x - 3 + 3 = 2x + 3
3x = 2x + 3

Subtract 2x on both sides

3x - 2x = 2x + 3 - 2x
3x - 2x = 3
x = 3

So, your final answer is x = 3.

tino4ka555 [31]2 years ago

5 0

$3x-3=2x \\ \\ -3 = 3x - 3x \ / \ subtract \ 3x \ from \ each \ side \\ \\ -3 = -x \ / \ simplify \\ \\ 3 = x \ / \ multiply \ each \ side \ by \ -1 \\ \\ x = 3 \ / \ switch \ the \ sides \\ \\ Answer: \fbox {x = 3}$

You might be interested in

What is the answer to 41x72.94

Alex Ar [27]

Answer:

2990.54

Step-by-step explanation:

The answer would be 2990.54.

3 0

2 years ago

In the following questions, look at different ways to represent the relation given by the equation y = x2 - 1. The table below s

Nady [450]

Answer:

a= 3

b= 0

Step-by-step explanation:

4 0

1 year ago

Suppose the radius of a circle is 3 units what is the circumference

HACTEHA [7]

Answer:

$6\pi$ or approximately $18.85$ units

Step-by-step explanation:

The formula for finding the circumference of a circle is $C=2\pi r$ , where $C$ = circumference and $r$ = radius. In this case, $r=3$ , so after plugging that into $C=2\pi r$ , we get:

$C=2\pi r\\ = 2*\pi *3\\=6\pi$

Since $\pi$ is approximately $3.1415$ , $6\pi =6*3.1415=18.849$ , which rounds to $18.85$ . Therefore, the circumference of the circle is $6\pi$ , or approximately $18.85$ units. Hope this helps!

8 0

2 years ago

Invested $8200 at the rate of 4.5% p.a. It earned $738 simple interest. The

Igoryamba

Answer:

2 yrs

Step-by-step explanation:

SI =PTR/100

738=8200 x T x 4.5 /100

738 x 100 / 8200 x 4.5 = T

T =2

3 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Cint%20%5Cfrac%7Bdx%7D%7Bxln%5E%7Bp%7Dx%20%7D" id="TexFormula1" title=" \int \frac{dx}{xl

Eva8 [605]

Substitute $u=\ln x$ , so that $\mathrm du=\dfrac{\mathrm dx}x$ . The integral is then equivalent to

$\displaystyle\int\frac{\mathrm dx}{x\ln^px}=\int\frac{\mathrm du}{u^p}=\begin{cases}\dfrac{u^{p+1}}{p+1}+C&\text{for }p\neq1\\\\\ln|u|+C&\text{for }p=1\end{cases}$

Then transforming back to $x$ gives

$\displaystyle\int\frac{\mathrm dx}{x\ln^px}=\begin{cases}\dfrac{\ln^{p+1}x}{p+1}+C&\text{for }p\neq1\\\\\ln|\ln x|+C&\text{for }p=1\end{cases}$

4 0

3 years ago