https://r2.erweima.ai/mathbot/image/17253128152515zm1qztl.jpg

resuelve solamente los ejercicios 95, 96, 99, 100, 101 dame todo el procedimiento posible y explica como lo hiciste 

I'll start by solving the problems you asked for (95, 96, 99, 100, and 101) from the image you provided. I'll provide detailed explanations and the steps for each.

### Problem 95: \( (1.302 \times 10^{-5}) \div (3.261 \times 10^{-2}) \)

**Solution:**
- To divide numbers in scientific notation, divide the coefficients and subtract the exponents:
\[
\frac{1.302 \times 10^{-5}}{3.261 \times 10^{-2}} = \frac{1.302}{3.261} \times 10^{-5 - (-2)}
\]
- First, calculate the division of the coefficients:
\[
\frac{1.302}{3.261} \approx 0.399
\]
- Subtract the exponents:
\[
10^{-5 - (-2)} = 10^{-5 + 2} = 10^{-3}
\]
- Combine the results:
\[
0.399 \times 10^{-3} = 3.99 \times 10^{-4}
\]
Thus, the result is \( 3.99 \times 10^{-4} \).

### Problem 96: \( \sqrt{8} \times \sqrt{2} \div \sqrt{4} \)

**Solution:**
- Use the property of square roots that states \( \sqrt{a} \times \sqrt{b} = \sqrt{a \times b} \):
\[
\sqrt{8} \times \sqrt{2} = \sqrt{8 \times 2} = \sqrt{16}
\]
- Simplify:
\[
\sqrt{16} = 4
\]
- Now, divide by \( \sqrt{4} \):
\[
4 \div \sqrt{4} = 4 \div 2 = 2
\]
Thus, the result is \( 2 \).

### Problem 99: Número de moléculas
This problem seems to involve the calculation of the number of molecules in a given sample, probably using Avogadro's number.

**Solution:**
- Let's assume that you need to calculate the number of molecules in a sample of a given amount (e.g., grams) using the molar mass and Avogadro's number \( N_A \).
- The formula for the number of molecules \( N \) is:
\[
N = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \times N_A
\]
If the problem provided specific data, such as a given mass or molar mass, you would substitute those values here.

### Problem 100: ¿A qué distancia puede usted ver?
This problem is likely related to the distance at which a person can see the horizon from a certain height.

**Solution:**
- The formula to calculate the distance to the horizon from a height \( h \) is:
\[
d = \sqrt{2Rh}
\]
where:
  - \( R \) is the radius of the Earth (\( \approx 6371 \) km),
  - \( h \) is the height above the Earth's surface.
  
If height \( h \) is given in meters, convert it to kilometers before substituting into the formula.

### Problem 101: Rapidez de un auto que patina
This problem involves calculating the speed of a skidding car using the friction coefficient.

**Solution:**
- The formula to find the speed \( v \) of a skidding car based on the friction coefficient \( \mu \) and the distance \( d \) it skidded is:
\[
v = \sqrt{2 \mu g d}
\]
where:
  - \( \mu \) is the coefficient of friction,
  - \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)),
  - \( d \) is the distance in meters.

Substitute the provided values for \( \mu \) and \( d \) into the formula to find \( v \).

---

If you need further details or clarification on any of these solutions, feel free to ask! Would you like to go over anything else?

**Here are some related questions you might find helpful:**

1. How do you perform operations with scientific notation?
2. What are the properties of square roots and how do they simplify expressions?
3. How can Avogadro's number be used in different types of chemical calculations?
4. How does the height of an observer affect the distance to the horizon?
5. How is the speed of a vehicle related to the coefficient of friction during skidding?

**Tip:** Always keep track of units when performing calculations, especially in problems involving scientific notation or physical constants!

Explore detailed solutions for mathematical problems 95, 96, 99, 100, and 101. Includes explanations on scientific notation division, properties of square roots, calculating molecules using Avogadro's number, distance to the horizon formula, and speed of a skidding car formula.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation