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3 years ago

11

The school carnival is divided into sections. The dunk tanks are in 1/10 of the carnival. Games are in 4/10 of the carnival. Stu

dent exhibits are in 5/10 of the carnival. What fraction of the carnival is dunk tanks and games?

2 answers:

Stels [109]3 years ago

5 0

The answer is 5/10 but when you simplify it will be 1/2

Setler79 [48]3 years ago

4 0

You could add the 1/10 and the 4/10 because you already know that 5/10 is the student exhibits. 1/10+4/10=5/10.

Since... yea, 5/10.

Or another way— just subtract 5/10(student exhibits) from the whole 10/10

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De un pliego rectangular de foami que mide 96 cm de largo por 72 cm de ancho,se requiere cortar cuadros de la mayor superficie p

kykrilka [37]

I do not understand ur language I speak English very well sorry I can't help you with your homework

7 0

2 years ago

5) In a urea manufacturing plan, all employees work a basic week of 40 hours. Ay overtime worked during weekdays is paid for at

Mkey [24]

Answer: $‭1,135.15‬

Step-by-step explanation:

First calculate his basic pay;

= 40 * 14.60

= $584

Total for overtime worked during the week;

= 15 * 14.60 * 1.25

= $‭273.75‬

Total for overtime worked on Saturday;

= 6 * 14.60 * 1.5

= $131.40

Total for overtime worked on Sunday;

= 5 * 14.60 * 2

= $146

Sum all the figures;

= 584 + 273.75 + 131.40 + 146

= $‭1,135.15‬

3 0

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

Can anyone help me with number 10??

Galina-37 [17]

Answer:

y=−4/3x+10

Step-by-step explanation:

Find the negative reciprocal of the slope of the original line and use the slope-intercept form y=mx+b to find the line perpendicular to 3x−4y=−2.

4 0

3 years ago

Which explicit formula best describes this sequence? (26, 32, 38, 44, 50, . . . )

Dominik [7]

The answer:

f(n) = 26+(n-1)

3 0

3 years ago

Read 2 more answers