I need help I can’t seem to figure it out

What is the value of x to the nearest tenth? A)7.2 B)4.8 C)12.0 D)9.2

Anika [276]

Answer:

x = 7.2

Step-by-step explanation:

a² + b² = c²

x² + (19.2/2)² = (24/2)²

x² + 9.6² = 12²

Subtract 9.6² from both sides

x² = 12² - 9.6²

x² = 51.84

x = 7.2

6 0

2 years ago

I need help ASAP like I might actually fail

natulia [17]

2x + 7 ≤ 3x - 5

Start by keeping the variable x on the left side of the inequality, and move everything else to the right side.

We can do this by subtracting 3x from both sides and subtracting 7 from both sides.

-x ≤ -12

Divide both sides by -1 to make x a positive; when dividing by a negative number you will flip the inequality sign, so ≤ becomes ≥.

x ≥ 12

Possible solutions for x are any number equal to or greater than 12.

8 0

2 years ago

2/4 + 3/9 in Modeling Fractions?

horrorfan [7]

Answer: 13/12

Step-by-step explanation:

Reduce the fraction 3/9 to the lowest terms by extracting and canceling out 3.

The least common multiple of 4 and 3 is 12. Convert 3/4 and 1/3 to fractions with denominator 12.

Since 9/12 and 4/12 have the same denominator, add them by adding their numerators.

Add 9 and 4 to get 13.

5 0

2 years ago

HELPPP

lana [24]

Answer:

so what are we trying to figure out?

Step-by-step explanation:

4 0

2 years ago

Find x? 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