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# Algebra

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1. The mass of Earth is about 6 x 10^21 metric tons. The mass of the sun is about 1.998 x 10^27 metric tons. About how many times the mass of Earth is the mass of the sun? Express the answer in scientific notation.

2. The perimeter P of a square of side x is given by the polynomial equation: P=4x. A baseball diamond is a square 90 ft on a side. Find the perimeter of a baseball diamond.

3. Hadley Electronics is marketing a new kind of plasma TV. The firm determines that when it sells x TVs, its total revenue R (the total amount of money taken in) will be R = 280x - 0.4x^2 dollars. What is the total revenue from the sale of 75 TVs?

4. The polynomial equation C = 0.041h - 0.018A - 2.69 can be used to estimate the lung capacity C, in liters, of a female of height h, in centimeters, and age A, in years. Find the lung capacity of a 20-year-old woman who is 165 cm tall.

5. A researcher wants to investigate the potential spread of germs by contact. She knows that the number of possible handshakes within a group of x people, assuming each person shakes every other person's hand only once, is given by the following formula.

Use this formula for the following exercises. N = (1/2)(x^2 - x).

a. There are 100 people at a party. How many handshakes are possible?

b. Everyone at a meeting shook hands with each other. There were 300 handshakes in all. How many people were at the meeting?

https://brainmass.com/math/basic-algebra/introductory-intermediate-algebra-problems-291543

#### Solution Preview

Please refer to the attachment for the solutions.

1. Earth vs. Sun. The mass of Earth is about 6 x 1021 metric tons. The mass of the sun is about 1.998 x 1027 metric tons. About how many times the mass of Earth is the mass of the sun? Express the answer in scientific notation.

Solution:

Mass of Sun/ Mass of Earth = 1.998 x 1027 metric tons/6 x 1021 metric tons = 333000
= 3.33 x 105.
So, the mass of the Sun is about 3.33 x 105 times the mass of the Earth.

2. Perimeter of a Baseball Diamond. The ...

#### Solution Summary

This solution is comprised of detailed explanation and step-by-step calculation of the given algebra problems and provides a clear perspective of the underlying concepts.

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