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3 years ago

What would be the actual length, if the length of the model is 25 ft and the scale factor is less

Mathematics

1 answer:

Sidana [21]3 years ago

6 0

Answer:

20ft

Explanation:

You multiply the model length by the scale factor to find the actual length. <u>If the scale factor is less than zero, then the actual size will be smaller than the model</u>. Therefore, your answer would be 20ft.

Specifically, the scale factor is .8 if the actual size is 20ft, 1 if the actual size is 25ft, and 1.2 if the actual size is 30ft.

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Step-by-step explanation:

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2 years ago

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Identify the y-intercept, constant factor, and exponent for each function below.

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2 years ago

1.1. Which statement explains why the two systems of equations below have the

marin [14]

Answer:

B. The first equation in B is the sum of the equations in A, while the other is obtained

by multiplying the first equation in A by 2.

Step-by-step explanation:

If we add the first two equations that shown in A so

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3 years ago

Compared to last year, the population of boom town has increased by 24%. The population is now 6,600. What was the population la

CaHeK987 [17]

Answer:

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8 0

3 years ago

Int(1 \(1 + {e}^{x} )

Andreyy89

Answer:

$\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx}= x - \ln(1 + e^{x}) + C\end{aligned}$ .

Step-by-step explanation:

The first derivative of the denominator $1 + e^{x}$ is $e^{x}$ . Rewrite the fraction to obtain that expression on the numerator.

$\begin{aligned}\frac{1}{1 + e^{x}} &= \frac{1 + e^{x}}{1 + e^{x}} - \frac{e^{x}}{1+e^{x}}\\&=1-\frac{e^{x}}{1+e^{x}}\end{aligned}$ .

In other words,

$\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx} &= \int{dx} - \int{\frac{e^{x}}{1+e^{x}}\cdot dx}\end{aligned}$ .

Apply $u$ -substitution on the integral $\displaystyle \int{\frac{e^{x}}{1+e^{x}}\cdot dx}$ :

Let $u = 1 + e^{x}$ . $u > 1$ .

$du = e^{x}\cdot dx$ .

$\displaystyle \int{\frac{e^{x}}{1+e^{x}}\cdot dx} = \int{\frac{du}{u}} = \ln{|u|} = \ln{u} +C = \ln{(1+e^{x})}+C$ .

Therefore

$\begin{aligned}\int{\frac{1}{1 + e^{x}}\cdot dx} &= \int{dx} - \int{\frac{e^{x}}{1+e^{x}}\cdot dx}\\ & = x - \ln{(1 + e^{x})}+C\end{aligned}$ .

7 0

2 years ago