Algebra 2 problems???

Question

Please help me! Please explain as well...thank u so much if u do help me.

1) the current in a stream moves at a speed of 8mph. A boat travels 34 mi upstream abd 34 mi downstream in a total time of 7hrs. WHat is the speed of the boat in still water?

2) If a firm invests amount Po( the o is a zero) at an interest rate of r%, then a year later invests amount P1 at the same rate, then the total amount at the end of two years is given bby the formula A=Po (1 + r)^2 + P1 (1 +r). A firm invests $4000, then a year later invests an additional $2200. At the end of two years, it has a total of $6720. What is the interest rate?
(Answer should be like this: The interest rate is r=___%)(plz, round to the nearest integer).

THANK U SO MUCH IF U HELP ME!!! U guys r my heros in math! :P

Answer 1

1) Let s be the speed of the boat.

s-8 equals the speed going upstream

s+8 equals the speed going downstream

The boat traveled 34 miles upstream and 34 miles downstream.

It took 7 hours to do this.

time = distance / speed

7 = 34/(s-8) + 34/(s+8)

Multiply both sides by (s+8) and (s-8)

7(s+8)(s-8) = 34(s+8) + 34(s-8)

Multiply it out

7s^2 - 448 = 68s

Subtract 68s from both sides

7s^2 - 68s - 448 = 0

Using the quadratic formula:

x = [ -b +/- sqrt(b^2 - 4ac) ] / 2a

s = [ 68 +/- sqrt(68^2 - (4)(7)(-448) ) ] / 14
s = 14.22 miles per hour
OR
s= -4.5 (but speed is always positive, so we can get rid of this one)

So, the speed is 14.22 miles per hour.

2) So, we know A=Po(1+r)^2 + P1(1+r)

Po = 4000
P1 = 2200
A = 6720

6720 = 4000(1 + r)^2 + 2200(1 + r)

Multiply out (1 + r)^2

6720 = 4000(1 + 2r + r^2) + 2200(1 + r)

Multiply out the 4000 and the 2200

6720 = 4000 + 8000r + 4000r^2 + 2200 + 2200r

Combine like terms

6720 = 4000r^2 + 10200r + 6200

Subtract 6720 from both sides

4000r^2 + 10200r - 520 = 0

Use the Quadratic Formula

r = [ -10200 +/- sqrt(10200^2 - (4)(4000)(-520)) ] / (2)(4000)

r = .05
OR
r = -2.6 (which is obviously not the right one)

so r = .05, which in percentage is

.05 * 100 = 5%

Answer 2

Hi,

1) the current in a stream moves at a speed of 8mph. A boat travels 34 mi upstream abd 34 mi downstream in a total time of 7hrs. WHat is the speed of the boat in still water?

upstream rate = 8 - x
downstream rate = 8 + x

distance/rate = time

34/(8 - x) + 34/(8 + x) = 7

Multiply each term by (8 - x)(8 + x) to eliminate fractions.

34(8 + x) + 34(8 - x) = 7(8 - x)(8 + x)

272 + 34x + 272 - 34x = 448 - 112x - 7x²







2) If a firm invests amount Po( the o is a zero) at an interest rate of r%, then a year later invests amount P1 at the same rate, then the total amount at the end of two years is given bby the formula A=Po (1 + r)^2 + P1 (1 +r). A firm invests $4000, then a year later invests an additional $2200. At the end of two years, it has a total of $6720. What is the interest rate?
(Answer should be like this: The interest rate is r=___%)(plz, round to the nearest integer).

Answer 3

1) The current in a stream moves at a speed of 8mph. A boat travels 34 mi upstream abd 34 mi downstream in a total time of 7hrs. What is the speed of the boat in still water?

Let
r = rate of boat in still water
r + 8 = speed downstream
r - 8 = speed upstream

This problem is of the form

time = distance / rate

7 = 34/(r + 8) + 34/(r - 8)

Multiply thru by (r + 8)(r - 8) to clear the denominators.

7(r + 8)(r - 8) = 34(r + 8) + 34(r - 8)
7(r² - 64) = 34r + 34*8 + 34r - 34*8
7r² - 448 = 68r
7r² - 68r - 448 = 0

r = [68 ± √[68² - 4*7*(-448)] / (2*7)
r = (68 ± √17168) / 14

r must be positive so the negative solution is eliminated.

r = (68 + √17168) / 14 ≈ 14.216194 mph

Answer 4

1) Let s be the speed in still water
There are two eqns:
34 = t(s-8) time t upstream 34 miles
t = 34/(s-8)
34 = (7-t)(s+8) time 7-t downstream 34 miles
Substitute the time t (such that t + 7-t = 7) in the second eqn:
34 = [7 - 34/(s-8)](s+8) = [7(s-8) - 34](s+8)/(s-8)
34(s-8) = (7s - 56 - 34)(s+8) = (7s-90)(s+8)
34s - 272 = 7s^2 + 56s - 90s - 720
34s - 272 = 7s^2 + 34s - 720
7s^2 = 720+272 = 992
s = sqr[992/7] = 11.9 mi/hour

2) Let x = 1+r
A = P0 x^2 + P1 x
P0 = $4000
P1 = $2200
A = $6720
P0 x^2 + P1 x - A = 0
x = {-P1 +- sqr[P1^2 - 4(P0)(-A)]}/2(P0)
= {-2200 +- sqr[2200^2 + 4(4000)(6720)]}/2(4000)
= { -2200 +- 10600}/8000
= (8400)/8000 = 21/20
1+r = 21/20
r = 21/20 - 1 = 21/20 - 20/20 = 1/20 = .05 = 5%