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papen · 10-31-2007, 02:02 PM

Can someone please help me with these math Questions..10 pts to correct answer! (please show work) 1.) A student invests $8,500 in a savings account drawing interest that is compounded annually. Find the annual rate if the money grows to $9,193.60 in 2 years 2.) The sum of the squares of two consecutive postive integers is 85. Find the integers. 3.) Find a third-degree equation that has a solution set of {2,3,-4}

ivory · 10-31-2007, 02:02 PM

2) 6 and 7 (36 + 49 = 85)

Marush · 10-31-2007, 02:02 PM

1. 8500 x (1+x)^2= 9163.60; rearrange and solve for x

2. 6 & 7 by inspection, 9^2 = 81, so you have an upper limit, start downwards from there.

3. how dumb are you?
(x-2)(x-3)(x+4)=0 multiply out

quicken · 10-31-2007, 02:02 PM

For the first problem you need the equation for compound interest. Here is a link. All you need to do is divided A by P
then take the square root and subtract 1 from both sides. Multiply both sides by n and you have solved for r. I am sure you can do the math for yourself.
For the second problem you can guess the answer fairly fast it has to be two numbers less than 10 since 10^2 is 100.
9^2=81,8^=64,7^=49,6^2=36 49+36=85 so it is 6 and 7.
For the third problem.
The solution set {2,3,-4} is the tidy up version of (x-2)=0,(x-3)=0, and (x+4)=0
so to find the equation multiply
(x-2)*(x-3)*(x+4)

Good Luck

Wastedspace · 10-31-2007, 02:02 PM

2) the two integers are 6 & 7

6^2 = 36, 7^2 = 49. 36+49 = 85

3) (x-2)(x-3)(x+4) = 0
multiply & u will get ur equation

vix · 10-31-2007, 02:02 PM

1. Let annual rate of interest be r%. After 1 year, the money would be worth $8 500*(1+r/100). After another year this new principal would be worth $8 500*(1+r/100)^2. So we need to solve
8 500(1+r/100)^2 = 9 193.6
(1+0.01r)^2 = 9193.6/8 500
(1+0.01r) = sqrt(1.0816) =1.04
so r = 0.04/0.01 = 4% is the annual rate of interest.

2. Let the two consecutive positive integers be x and x+1.
We need to solve x^2 + (x+1)^2 =85. Expanding (x+1)^2 and simplifying gives
x^2+x^2+2x+1 =85
2x^2+2x+1-85 = 0
2(x^2+x-42) = 0
Factorising this quadratic gives
2(x+7)(x-6) = 0
which has solutions x=6 or -7. Since we're asked for a positive integer we take x=6.

Check: 6^2 + 7^2= 36 +49 =85.

3. This is saying that the third-degree equation has factors (x-2), (x-3) and (x+4), so just multiply these together (expand) and equate to zero to get the required equation

(x-2)(x-3)(x+4) = 0
[x^2-2x-3x+6](x+4) = 0
[x^2-5x+6](x+4) = 0
x^3+4x^2-5x^2-20x+6x+24 = 0
x^3-x^2-14x+24 = 0
is the required third degree equation.

10-31-2007, 02:02 PM	#1
papen Greenhorn Join Date: Jul 2006 Posts: 21 Rep Power: 6	Math question...10 pts to correct answer? Can someone please help me with these math Questions..10 pts to correct answer! (please show work)<br /><br />1.) A student invests $8,500 in a savings account drawing interest that is compounded annually. Find the annual rate if the money grows to $9,193.60 in 2 years<br /><br />2.) The sum of the squares of two consecutive postive integers is 85. Find the integers.<br /><br />3.) Find a third-degree equation that has a solution set of <br />{2,3,-4}<br />

10-31-2007, 02:02 PM	#2
ivory Greenhorn Join Date: Jul 2006 Posts: 10 Rep Power: 0	Math question...10 pts to correct answer? 2) 6 and 7 (36 + 49 = 85)

10-31-2007, 02:02 PM	#3
Marush Greenhorn Join Date: Jul 2006 Posts: 22 Rep Power: 6	Math question...10 pts to correct answer? 1. 8500 x (1+x)^2= 9163.60; rearrange and solve for x 2. 6 & 7 by inspection, 9^2 = 81, so you have an upper limit, start downwards from there. 3. how dumb are you? (x-2)(x-3)(x+4)=0 multiply out

10-31-2007, 02:02 PM	#4
quicken Greenhorn Join Date: Jul 2006 Posts: 27 Rep Power: 6	Math question...10 pts to correct answer? For the first problem you need the equation for compound interest. Here is a link. All you need to do is divided A by P then take the square root and subtract 1 from both sides. Multiply both sides by n and you have solved for r. I am sure you can do the math for yourself. For the second problem you can guess the answer fairly fast it has to be two numbers less than 10 since 10^2 is 100. 9^2=81,8^=64,7^=49,6^2=36 49+36=85 so it is 6 and 7. For the third problem. The solution set {2,3,-4} is the tidy up version of (x-2)=0,(x-3)=0, and (x+4)=0 so to find the equation multiply (x-2)(x-3)(x+4) Good Luck

10-31-2007, 02:02 PM	#5
Wastedspace Greenhorn Join Date: Jul 2007 Posts: 22 Rep Power: 5	Math question...10 pts to correct answer? 2) the two integers are 6 & 7 6^2 = 36, 7^2 = 49. 36+49 = 85 3) (x-2)(x-3)(x+4) = 0 multiply & u will get ur equation

10-31-2007, 02:02 PM	#6
vix Greenhorn Join Date: Jul 2006 Posts: 9 Rep Power: 0	Math question...10 pts to correct answer? 1. Let annual rate of interest be r%. After 1 year, the money would be worth $8 500(1+r/100). After another year this new principal would be worth $8 500(1+r/100)^2. So we need to solve 8 500(1+r/100)^2 = 9 193.6 (1+0.01r)^2 = 9193.6/8 500 (1+0.01r) = sqrt(1.0816) =1.04 so r = 0.04/0.01 = 4% is the annual rate of interest. 2. Let the two consecutive positive integers be x and x+1. We need to solve x^2 + (x+1)^2 =85. Expanding (x+1)^2 and simplifying gives x^2+x^2+2x+1 =85 2x^2+2x+1-85 = 0 2(x^2+x-42) = 0 Factorising this quadratic gives 2(x+7)(x-6) = 0 which has solutions x=6 or -7. Since we're asked for a positive integer we take x=6. Check: 6^2 + 7^2= 36 +49 =85. 3. This is saying that the third-degree equation has factors (x-2), (x-3) and (x+4), so just multiply these together (expand) and equate to zero to get the required equation (x-2)(x-3)(x+4) = 0 [x^2-2x-3x+6](x+4) = 0 [x^2-5x+6](x+4) = 0 x^3+4x^2-5x^2-20x+6x+24 = 0 x^3-x^2-14x+24 = 0 is the required third degree equation.

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