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3 years ago

PLEASE HELP!!! BRAINLIEST IF CORRECT! 40 POINTS

2 answers:

7 0

1. D; -5m+2m, because -5+2=-3

2. B; 4 terms, every part of an expression that's not the subtraction/addition sign counts as a term.

3. Not 100% sure about this one, but I'll go with B; 4x².

4. C; 5b, that term cannot be combined because the variable is different from the rest.

I hope this was helpful.

PIT_PIT [208]3 years ago

6 0

One
The five has to be minus, the two is plus. m has to be both. That makes D the answer. The others give 3m 7m and 3m.
Answer D

Two
This question is VERY important. Don;t leave it alone until you understand it.
2x and x can combine. Nothing else can combine with anything else. The answer you get is

3x - 5y + 3

There are 3 different terms. Nothing else will combine because either the letters are different (x and y) or the 3 has no letter with it. This is your answer. Algebra is real picky about that.

Answer A

Three
Another biggy. x^2 is not the same thing as x. The other three will combine by addition. x^2 must be left alone along with 4.
Answer B 4x^2
If you were actually to do this, you would get 9x + 4x^2
Four
5b. the letters are not the same.

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2 years ago

Find x?<br> In 3x - In(x - 4) = ln(2x - 1) +ln3

earnstyle [38]

Answer:

$x = \displaystyle \frac{5 + \sqrt{17}}{2}$ .

Step-by-step explanation:

Because $3\, x$ is found in the input to a logarithm function in the original equation, it must be true that $3\, x > 0$ . Therefore, $x > 0$ .

Similarly, because $(x - 4)$ and $(2\, x - 1)$ are two other inputs to the logarithm function in the original equation, they should also be positive. Therefore, $x > 4$ .

Let $a$ and $b$ represent two positive numbers (that is: $a > 0$ and $b > 0$ .) The following are two properties of logarithm:

$\displaystyle \ln (a) + \ln(b) = \ln\left(a \cdot b\right)$ .

$\displaystyle \ln (a) - \ln(b) = \ln\left(\frac{a}{b}\right)$ .

Apply these two properties to rewrite the original equation.

Left-hand side of this equation:

$\begin{aligned}&\ln(3\, x) - \ln(x - 4)= \ln\left(\frac{3\, x}{x -4}\right)\end{aligned}$

Right-hand side of this equation:

$\ln(2\, x- 1) + \ln(3) = \ln\left(3 \left(2\, x - 1\right)\right)$ .

Equate these two expressions:

$\begin{aligned}\ln\left(\frac{3\, x}{x -4}\right) = \ln(3(2\, x - 1))\end{aligned}$ .

The natural logarithm function $\ln$ is one-to-one for all positive inputs. Therefore, for the equality $\begin{aligned}\ln\left(\frac{3\, x}{x -4}\right) = \ln(3(2\, x - 1))\end{aligned}$ to hold, the two inputs to the logarithm function have to be equal and positive. That is:

$\displaystyle \frac{3\ x}{x - 4} = 3\, (2\, x - 1)$ .

Simplify and solve this equation for $x$ :

$x^2 - 5\, x + 2 = 0$ .

There are two real (but not rational) solutions to this quadratic equation: $\displaystyle \frac{5 + \sqrt{17}}{2}$ and $\displaystyle \frac{5 - \sqrt{17}}{2}$ .

However, the second solution, $\displaystyle \frac{5 - \sqrt{17}}{2}$ , is not suitable. The reason is that if $x = \displaystyle \frac{5 - \sqrt{17}}{2}$ , then $(x - 4)$ , one of the inputs to the logarithm function in the original equation, would be smaller than zero. That is not acceptable because the inputs to logarithm functions should be greater than zero.

The only solution that satisfies the requirements would be $\displaystyle \frac{5 + \sqrt{17}}{2}$ .

Therefore, $x = \displaystyle \frac{5 + \sqrt{17}}{2}$ .

7 0

2 years ago