1) John usually gets to work by driving 29 miles down Greenwich Street, which goes from his house to his office in a straight line. One week, Greenwich Street is closed due to construction, so John gets to work by driving 13 miles wests on 57th street, and then turning onto Harrison Street and driving north to his office. Approximately how many more miles does John have to drive by taking the detour?
^18 miles
2) From 1980 to 2015, the student body at University A grew approximately 3% per year. By contrast, during the same period, the student body at University B grew by approxiametely 300 students per year. If both universities had exactly 10,000 students in 1980, what was the difference between the student bodies of University A and University B 2015?
^464
3) What is one possible solution to the equation
6/(x+1) minus 3/(x-1) equals 1/4?
1. second leg=sqrt(29^2-13^2)=25.9
tripexcess=25.9+13-29= not your answer
2. A=10000*1.03^35=28136
b=10000+35*300=you do it.
How is number two possible?? 20,500? How?
Yes, University B will have 20,500 students. As Bob showed, University A will have 28,136 students. What your question asks for is the difference between them. What is that difference?
How did you arrive at the answer 464?
To calculate the answer to each question, follow these steps:
1) For the first question, you need to calculate the extra distance John has to drive by taking the detour. Start by finding the difference between the original route (29 miles down Greenwich Street) and the new route (13 miles west on 57th street and then north on Harrison Street).
- John's original route is 29 miles.
- John's new route is 13 miles west on 57th street and then turning north on Harrison Street.
To find the difference, subtract the new route distance from the original route distance:
- 29 miles - (13 miles + x miles) = 18 miles
Therefore, John has to drive an additional 18 miles by taking the detour.
2) For the second question, you need to calculate the difference between the student bodies of University A and University B in 2015.
- Both universities had 10,000 students in 1980.
- University A's student body grew approximately 3% per year.
- University B's student body grew by approximately 300 students per year.
To find the difference, calculate the student body for each university in 2015 and then subtract them:
- For University A in 2015, calculate: 10,000 * (1 + 0.03)^35 = approximately 24,229 students.
- For University B in 2015, calculate: 10,000 + 300 * 35 = 20,500 students.
Subtract University B's student count from University A's student count:
- 24,229 - 20,500 = 3,729
Therefore, the difference between the student bodies of University A and University B in 2015 is approximately 3,729 students.
3) For the third question, you need to find one possible solution to the equation:
6/(x+1) - 3/(x-1) = 1/4
To solve this equation, you can find a common denominator, simplify the equation, and solve for x. Follow these steps:
- The common denominator is (x+1)(x-1). Multiply each term by the common denominator to get rid of the fractions.
6(x-1) - 3(x+1) = 1(x+1)(x-1)/4
- Simplify and distribute:
6x - 6 - 3x - 3 = (x^2 - 1)/4
3x - 9 = (x^2 - 1)/4
- Multiply every term by 4 to get rid of the divisor:
12x - 36 = x^2 - 1
- Rearrange the equation and set it equal to zero by moving all terms to one side:
0 = x^2 - 12x + 35
- Factor the equation:
0 = (x - 5)(x - 7)
- Set each factor equal to zero:
x - 5 = 0 -> x = 5
x - 7 = 0 -> x = 7
Therefore, x can be either 5 or 7. These are two possible solutions to the equation.