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Po's dog weighs 24 and 65 lb Ricky's dog weighs 42 LB how much more does Ricky's dog weigh

arsen [322]

41 lb
work: 65 - 24 = 41
you're welcome

8 0

3 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

Integration of the following function

DerKrebs [107]

Answer: None of the above

Step-by-step explanation:

Given

$\frac{\mathrm{d} \bar{c}}{\mathrm{d} x}=-6x^{-2}+\frac{1}{6}$

$d\bar{c}=\left ( -6x^{-2}+\frac{1}{6}\right )dx$

Now, integrating both sides

$\int d\bar{c}=\int \left ( -6x^{-2}+\frac{1}{6}\right )dx$

$\bar{c}=-6\frac{x^{-2+1}}{-2+1}+\frac{1}{6}x+k_1$

$\bar{c}=6x^{-1}+\frac{1}{6}x+k_1$

6 0

2 years ago

X^2 - 10x - 20 = 0, which number would have to be added to "complete the square"?

Airida [17]

Answer:

45

Step-by-step explanation:

When you're completing the square you're adding (b/2)^2 which in this case is (-5)^2 or 25. This is because when you rewrite it as a square binomial you write it in the form (x+(b/2))^2 which is going to result in x^2+b+(b/2)^2. Since you need to reach 25 you need to add 45 since the last value is at -20.

3 0

1 year ago

Read 2 more answers

Select the expression that shows the application of the distributive property for

sergeinik [125]

Answer:

c. 9t+45 that shows distributive property

where it shows a*(b+c)=(a*b)+(b*c).so,9(t+5)=9*t+9*5

8 0

2 years ago

Ple help me pls help me​

Ple help me pls help me