Algebra ->  Tutoring on algebra.com -> See tutors' answers!      Log On

 Tutoring Home For Students Tools for Tutors Our Tutors Register Recently Solved
 By Tutor
| By Problem Number |

Tutor:

# Recent problems solved by 'nerdybill'

Jump to solutions: 0..29 , 30..59 , 60..89 , 90..119 , 120..149 , 150..179 , 180..209 , 210..239 , 240..269 , 270..299 , 300..329 , 330..359 , 360..389 , 390..419 , 420..449 , 450..479 , 480..509 , 510..539 , 540..569 , 570..599 , 600..629 , 630..659 , 660..689 , 690..719 , 720..749 , 750..779 , 780..809 , 810..839 , 840..869 , 870..899 , 900..929 , 930..959 , 960..989 , 990..1019 , 1020..1049 , 1050..1079 , 1080..1109 , 1110..1139 , 1140..1169 , 1170..1199 , 1200..1229 , 1230..1259 , 1260..1289 , 1290..1319 , 1320..1349 , 1350..1379 , 1380..1409 , 1410..1439 , 1440..1469 , 1470..1499 , 1500..1529 , 1530..1559 , 1560..1589 , 1590..1619 , 1620..1649 , 1650..1679 , 1680..1709 , 1710..1739 , 1740..1769 , 1770..1799 , 1800..1829 , 1830..1859 , 1860..1889 , 1890..1919 , 1920..1949 , 1950..1979 , 1980..2009 , 2010..2039 , 2040..2069 , 2070..2099 , 2100..2129 , 2130..2159 , 2160..2189 , 2190..2219 , 2220..2249 , 2250..2279 , 2280..2309 , 2310..2339 , 2340..2369 , 2370..2399 , 2400..2429 , 2430..2459 , 2460..2489 , 2490..2519 , 2520..2549 , 2550..2579 , 2580..2609 , 2610..2639 , 2640..2669 , 2670..2699 , 2700..2729 , 2730..2759 , 2760..2789 , 2790..2819 , 2820..2849 , 2850..2879 , 2880..2909 , 2910..2939 , 2940..2969 , 2970..2999 , 3000..3029 , 3030..3059 , 3060..3089 , 3090..3119 , 3120..3149 , 3150..3179 , 3180..3209 , 3210..3239 , 3240..3269 , 3270..3299 , 3300..3329 , 3330..3359 , 3360..3389 , 3390..3419 , 3420..3449 , 3450..3479 , 3480..3509 , 3510..3539 , 3540..3569 , 3570..3599 , 3600..3629 , 3630..3659 , 3660..3689 , 3690..3719 , 3720..3749 , 3750..3779 , 3780..3809 , 3810..3839 , 3840..3869 , 3870..3899 , 3900..3929 , 3930..3959 , 3960..3989 , 3990..4019 , 4020..4049 , 4050..4079 , 4080..4109 , 4110..4139 , 4140..4169 , 4170..4199 , 4200..4229 , 4230..4259 , 4260..4289 , 4290..4319 , 4320..4349 , 4350..4379 , 4380..4409 , 4410..4439 , 4440..4469 , 4470..4499 , 4500..4529 , 4530..4559 , 4560..4589 , 4590..4619 , 4620..4649 , 4650..4679 , 4680..4709 , 4710..4739 , 4740..4769 , 4770..4799 , 4800..4829 , 4830..4859 , 4860..4889 , 4890..4919 , 4920..4949 , 4950..4979 , 4980..5009 , 5010..5039 , 5040..5069 , 5070..5099 , 5100..5129 , 5130..5159 , 5160..5189 , 5190..5219 , 5220..5249 , 5250..5279 , 5280..5309 , 5310..5339 , 5340..5369 , 5370..5399 , 5400..5429 , 5430..5459 , 5460..5489 , 5490..5519 , 5520..5549 , 5550..5579 , 5580..5609 , 5610..5639 , 5640..5669 , 5670..5699 , 5700..5729 , 5730..5759 , 5760..5789 , 5790..5819 , 5820..5849 , 5850..5879 , 5880..5909 , 5910..5939 , 5940..5969 , 5970..5999 , 6000..6029 , 6030..6059 , 6060..6089 , 6090..6119 , 6120..6149 , 6150..6179 , 6180..6209 , 6210..6239 , 6240..6269 , 6270..6299 , 6300..6329 , 6330..6359 , 6360..6389 , 6390..6419 , 6420..6449 , 6450..6479 , 6480..6509 , 6510..6539 , 6540..6569 , 6570..6599 , 6600..6629 , 6630..6659 , 6660..6689 , 6690..6719 , 6720..6749 , 6750..6779 , 6780..6809 , 6810..6839 , 6840..6869 , 6870..6899 , 6900..6929 , 6930..6959 , 6960..6989 , 6990..7019 , 7020..7049 , 7050..7079 , 7080..7109, >>Next

Quadratic_Equations/194461: the function h=-5t2+20t+1 models the height, h meters, of a baseball as a function of the time, t seconds, since it was hit. The ball hit the ground before the fielder could catch it. Use the quadratic formula to solve the following problems.
a) How long was the baseball in the air, to the nearest tenth of a second?
b) For how many seconds was the height of the ball at least 16 m?
1 solutions

Answer 145906 by nerdybill(7090)   on 2009-05-03 17:03:13 (Show Source):
You can put this solution on YOUR website!
the function h=-5t2+20t+1 models the height, h meters, of a baseball as a function of the time, t seconds, since it was hit. The ball hit the ground before the fielder could catch it. Use the quadratic formula to solve the following problems.
a) How long was the baseball in the air, to the nearest tenth of a second?
.
Set h=0 and solve for t
h=-5t2+20t+1
0=-5t2+20t+1
Solving via the quadratic equation yields:
x ={-0.04939, 4.0494}
See below for details...
We can toss out the negative solution leaving:
x = 4.0494 seconds
.
b) For how many seconds was the height of the ball at least 16 m?
Set h=0 and solve for t
h=-5t2+20t+1
16=-5t2+20t+1
0=-5t2+20t-15
0 = (-5t+15)(t-1)
t = {1,3}
This means on the way up at 1 second it reaches 16m and then on the way down at 3 seconds it passes 16m again.
Thus it was in the air for 2 seconds.
.
 Solved by pluggable solver: SOLVE quadratic equation with variable Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=420 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: -0.0493901531919196, 4.04939015319192. Here's your graph:

 Geometry_Word_Problems/194396: The perimeter of a basketball court is 102 meters. The length of the court is 6 meters more than twice the width. What are the dimensions of the court?1 solutions Answer 145866 by nerdybill(7090)   on 2009-05-03 10:07:55 (Show Source): You can put this solution on YOUR website! The perimeter of a basketball court is 102 meters. The length of the court is 6 meters more than twice the width. What are the dimensions of the court? . Let w = width of court then 2w+6 = length of court . 2(w+ 2w+6) = 102 2w+ 4w+12 = 102 6w+12 = 102 6w = 90 w = 15 meters (width) . Length: 2w+6 = 2(15)+6 = 30+6 = 36 meters (length) . dimensions: 15x36 meters
 Square-cubic-other-roots/194293: This question is from textbook saxon algebra 2 If 6 square root 6x-1-18(-18 is not under square root sign)=0, then x= 5/6. Did I calculate this correctly?1 solutions Answer 145801 by nerdybill(7090)   on 2009-05-02 15:57:11 (Show Source): You can put this solution on YOUR website!Is this what you meant: . . . . . . .
 Roman-numerals/194299: Can someone help, please? I got -2n+4m. Thanks a lot Collect like terms 14m+5n-10m-7n= 4m+-2n1 solutions Answer 145800 by nerdybill(7090)   on 2009-05-02 15:52:31 (Show Source): You can put this solution on YOUR website!You are correct. However, instead of: -2n+4m write: 4m-2n (looks better) . Also, you could factor out a 2: 2(2m-n)
 Systems-of-equations/194281: Solve equation for x. 1/27=3^2x1 solutions Answer 145795 by nerdybill(7090)   on 2009-05-02 14:34:45 (Show Source): You can put this solution on YOUR website!1/27=3^2x Take log3 (log base 3) of both sides: log3(1/27) = x log3(1) - log(327) = x log3(3^0) - log3(3^3) = x 0-3 = x -3 = x
Square-cubic-other-roots/194265: How do I solve this eqaution by completing the sqaure?
5x^2-10x+2=0
1 solutions

Answer 145794 by nerdybill(7090)   on 2009-05-02 13:43:39 (Show Source):
You can put this solution on YOUR website!
5x^2-10x+2=0
Isolate the x-terms on the left:
5x^2-10x = -2
Divide both sides by 5:
x^2-2x = -2/5
Now, we can complete the square on the left:
(x^2-2x+___) = -2/5
(x^2-2x+1) = -2/5+1
(x-1)^2 = -2/5 + 5/5
(x-1)^2 = 3/5
Take the square root of both sides:
x-1 = sqrt(3/5) = sqrt(15)/5
x = sqrt(15)/5 + 1
x = 1 +- sqrt(15)/5
or
x = (5 +- sqrt(15))/5
.
To check, we can solve it using the quadratic equation which yields:
x = {1.7746, 0.2254}
Which gives you the same answers.
 Solved by pluggable solver: SOLVE quadratic equation with variable Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=60 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 1.77459666924148, 0.225403330758517. Here's your graph:

 logarithm/194257: Solve: log(6)(x+7)-log(6)(x-2)=log(6)51 solutions Answer 145789 by nerdybill(7090)   on 2009-05-02 12:02:26 (Show Source): You can put this solution on YOUR website!Apply "log rules": log(6)(x+7)-log(6)(x-2)=log(6)5 log(6)((x+7)/(x-2))=log(6)5 (x+7)/(x-2) = 5 (x+7) = 5(x-2) x+7 = 5x-10 7 = 4x-10 17 = 4x 17/4 = x
 Geometry_Word_Problems/194227: Hi!!! Im Very Confused with this problem: The Longest side of a triangle is twice as long as the shortest side and 2cm longer than the third side. if the perimeter of the triangle is 33 cm, what is the lenght of each side? Pls Help me!!! Thank you!!!1 solutions Answer 145770 by nerdybill(7090)   on 2009-05-02 04:50:48 (Show Source): You can put this solution on YOUR website!The Longest side of a triangle is twice as long as the shortest side and 2cm longer than the third side. if the perimeter of the triangle is 33 cm, what is the lenght of each side? . Let x = length of the longest side y = length of the third side z = length of the shortest side . then x = 2z x = y+2 x+y+z = 33 . Rearranging terms: x + 0y - 2z = 0 x - y + 0z = 2 x + y + z = 33 . Solving with Cramer's rule Matrix of coefficients: 1 0 -2 1 -1 0 1 1 1 Determinant of coefficients: (-1 - 0) - 2(1 + 1) = -1 -2(2) = -1 -4 = -5 . Matrix of 'x': 0 0 -2 2 -1 0 33 1 1 Determinant of 'x': -2(2+33) = -2(35) = -70 x = detx/detcoeff = -70/-5 = 14 cm . Matrix of 'y': 1 0 -2 1 2 0 1 33 1 Determinant of 'y': (2-0) - 2(33-2) = 2 - 2(31) = 2 -62 = -60 y = dety/detcoeff = -60/-5 = 12 cm . Matrix of 'z': 1 0 0 1 -1 2 1 1 33 Determinant of 'z': (-33-2) = -35 z = detz/detcoeff = -35/-5 = 7 cm . The three sides are: 14 cm, 12 cm and 7 cm
 logarithm/194232: log4 (x+3)=41 solutions Answer 145769 by nerdybill(7090)   on 2009-05-02 04:22:40 (Show Source): You can put this solution on YOUR website! log4 (x+3)=4 x+3 = 4^4 x+3 = 256 x = 256-3 x = 253
logarithm/194220: logx (2x + 5) = 2
1 solutions

Answer 145766 by nerdybill(7090)   on 2009-05-02 01:03:58 (Show Source):
You can put this solution on YOUR website!
logx (2x + 5) = 2
(2x + 5) = x^2
0 = x^2 - 2x - 5
.
Solving with the quadratic equation yields:
x = {3.449, -1.449}
We can toss out the negative solution leaving:
x = 3.449
.
.
 Solved by pluggable solver: SOLVE quadratic equation with variable Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=24 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 3.44948974278318, -1.44948974278318. Here's your graph:

 Finance/194218: Can you help me please? Solve the system by the addition method. x – y = 100 0.20x + 0.06y = 150 Thank you all1 solutions Answer 145765 by nerdybill(7090)   on 2009-05-02 00:59:02 (Show Source): You can put this solution on YOUR website!Starting with: x – y = 100 0.20x + 0.06y = 150 . Multiply both sides of the top equation by 0.06: 0.06x – 0.06y = 6 0.20x + 0.06y = 150 . Now, add both top and bottom equation together: 0.06x – 0.06y = 6 0.20x + 0.06y = 150 ---------------------- 0.26x = 156 x = 156/0.26 x = 600 . Substitute the above back into the first equation: x – y = 100 600 – y = 100 -y = 400 y = -400 . (x,y) = (600, -400)
 logarithm/194210: Simplify each of the following to a single logarithm: 1/2log(4x^2)+5log3x-log21 solutions Answer 145764 by nerdybill(7090)   on 2009-05-01 23:37:04 (Show Source): You can put this solution on YOUR website!1/2log(4x^2)+5log3x-log2 = log(4x^2)^(1/2) + log(3x)^5 - log 2 = log(2x) + log(243x^5) - log 2 = log(2x)(243x^5) - log 2 = log(486x^6) - log 2 = log(486x^6)/2 = log(243x^6)
 logarithm/194212: log x = log6 36 + log81 9 – log2 4 1 solutions Answer 145763 by nerdybill(7090)   on 2009-05-01 23:33:55 (Show Source): You can put this solution on YOUR website!log x = log6 36 + log81 9 – log2 4 we can rewrite as: log x = log6 6^2 + log81 81^(1/2) – log2 2^2 Now we simplify: log x = 2 + (1/2) – 2 log x = 1/2 x = 10^(1/2) x = 3.162
 logarithm/194205: Simplify each of the following to a single logarithm: (i) 2log 5x – log x 1 solutions Answer 145759 by nerdybill(7090)   on 2009-05-01 22:59:58 (Show Source): You can put this solution on YOUR website!2log 5x – log x . = log (5x)^2 – log x = log (25x^2) – log x = log (25x^2)/x = log (25x)
 Quadratic-relations-and-conic-sections/193913: Write an equation of a parabola with a vertex at the origin. 8. focus (0, 2) 9. focus (4, 0) 10. directrix at y = -81 solutions Answer 145729 by nerdybill(7090)   on 2009-05-01 14:05:49 (Show Source): You can put this solution on YOUR website! 8. focus (0, 2) This is a "vertical" parabola, if so, it follows: . Since, the vertex is at the origin: (h,k) = (0,0) 'c' is the distance from the vertex to the focus therefore c =2 Now, we plug it in to . 9. focus (4, 0) This is a "horizontal" parabola, if so, it follows: . Since, the vertex is at the origin: (h,k) = (0,0) 'c' is the distance from the vertex to the focus therefore c =4 . Now, we plug it in to . 10. directrix at y = -8 This is a "vertical" parabola, if so, it follows: . Since, the vertex is at the origin: (h,k) = (0,0) 'c' is the distance from the vertex to the directrix therefore c =-8 Now, we plug it in to
 Quadratic-relations-and-conic-sections/194144: This question is from textbook : how would you find the foci of this ellipse? x^2+9y^2=11 solutions Answer 145725 by nerdybill(7090)   on 2009-05-01 13:44:02 (Show Source): You can put this solution on YOUR website!. Equation for any ellipse: . So, we can rewrite: As: . From the above we now know: (h,k) = (0,0) a = 1 b = 1/3 . a^2 - b^2 = c^2 1^2 - (1/3)^2 = c^2 1 - 1/9 = c^2 8/9 = c^2 sqrt(8/9) = c (2/3)sqrt(2) = c . And, since 'a' is larger than 'b' -- it is horizontal major therefore, foci is at (h+-c, k) or (0+-(2/3)sqrt(2), 0 ) . Foci would then be: ((2/3)sqrt(2), 0 ) and (-(2/3)sqrt(2), 0 )
 Exponential-and-logarithmic-functions/193874: let log2=0.03+log3=0.05 find log(3/2)1 solutions Answer 145723 by nerdybill(7090)   on 2009-05-01 13:29:59 (Show Source): You can put this solution on YOUR website!Applying log rules: let log2=0.03+log3=0.05 find log(3/2) . log(3/2) can be rewritten as: log(3) - log(2) then, replace with given values: 0.05 - 0.03 = 0.02
 Travel_Word_Problems/194118: This morning, my dog Walter broke out of the house and ran directly to the farm up the road at a speed of 12mph. When he arrived at the farm he was scared by the cows so he turned around and ran directly home (by the same route) at a speed of 8mph. Walter spent a total time of one hour running to the farm and back. Find the distance from my house to the farm.1 solutions Answer 145698 by nerdybill(7090)   on 2009-05-01 09:52:14 (Show Source): You can put this solution on YOUR website!. Let t = time to get to the farm then 1-t = time to get back to the house . Applying the distance formula: d = rt where d is distance r is rate or speed t is time . 12t = 8(1-t) 12t = 8-8t 20t = 8 t = 8/20 t = 2/5 hour (to run from house to farm) . Distance then is: d = (2/5)12 d = 24/5 = 4.8 miles (distance from house to farm)
 logarithm/194059: express as an equivalent expression that is a single logarithm and if possible simplify: log(a)2x+3(log(a)x-log(a)y)1 solutions Answer 145697 by nerdybill(7090)   on 2009-05-01 09:44:36 (Show Source): You can put this solution on YOUR website!Apply "log rules". You can review them here: http://www.purplemath.com/modules/logrules.htm . log(a)2x+3(log(a)x-log(a)y) First, combine terms inside parenthesis: log(a)2x+3(log(a)(x/y)) Next, simplify right term: log(a)2x + log(a)(x/y)^3 Finally: log(a)(2x)(x/y)^3 log(a)(2x)(x^3/y^3) log(a)(2^4/y^3)
 logarithm/194114: express as an equivalent expression that is a single logarithm and if possible simplify. log(a)2x+3(log(a)x-log(a)y) 1 solutions Answer 145696 by nerdybill(7090)   on 2009-05-01 09:40:41 (Show Source): You can put this solution on YOUR website!Apply "log rules". You can review them here: http://www.purplemath.com/modules/logrules.htm . log(a)2x+3(log(a)x-log(a)y) First, combine terms inside parenthesis: log(a)2x+3(log(a)(x/y)) Next, simplify right term: log(a)2x + log(a)(x/y)^3 Finally: log(a)(2x)(x/y)^3 log(a)(2x)(x^3/y^3) log(a)(2^4/y^3)
 Polynomials-and-rational-expressions/194111: i have trouble factoring this out. please help. 3x^2-14x+8 thanks1 solutions Answer 145693 by nerdybill(7090)   on 2009-05-01 08:52:11 (Show Source): You can put this solution on YOUR website!3x^2-14x+8 (3x-2)(x-4)
 Exponential-and-logarithmic-functions/194054: What is log(1/10)?1 solutions Answer 145653 by nerdybill(7090)   on 2009-04-30 20:36:24 (Show Source): You can put this solution on YOUR website!Applying "log rules", we can rewrite: log(1/10) as = log(1) - log(10) = log(10^0) - log(10^1) = 0-1 = -1
 Travel_Word_Problems/193801: The profit on a watch is given by P = x^2 – 5x – 9 and where x is the number of watches sold per day. How many watches were sold on a day when there was a \$15 loss? 1 solutions Answer 145491 by nerdybill(7090)   on 2009-04-29 18:05:30 (Show Source): You can put this solution on YOUR website! The profit on a watch is given by P = x^2 – 5x – 9 and where x is the number of watches sold per day. How many watches were sold on a day when there was a \$15 loss? . Simply set P to -15 and solve for x: P = x^2 – 5x – 9 15 = x^2 – 5x – 9 0 = x^2 – 5x – 24 0 = (x-8)(x+3) x = {-3, 8} . Toss out the negative solution leaving: x = 8
 Miscellaneous_Word_Problems/193803: One sixth of all sales at Johns Bike Shop are for cash. If cash sales for the week were \$265.00. What were Johns total sales?1 solutions Answer 145489 by nerdybill(7090)   on 2009-04-29 18:03:07 (Show Source): You can put this solution on YOUR website! One sixth of all sales at Johns Bike Shop are for cash. If cash sales for the week were \$265.00. What were Johns total sales? . Let x = total sales then (1/6)x = 265 x/6 = 265 x = 265*6 x = \$1590
 Travel_Word_Problems/193806: A pair of cuff links was marked up \$225 from cost, which amounts to a 50% increase. Find the original cost of the pair of cuff links1 solutions Answer 145488 by nerdybill(7090)   on 2009-04-29 18:00:59 (Show Source): You can put this solution on YOUR website! A pair of cuff links was marked up \$225 from cost, which amounts to a 50% increase. Find the original cost of the pair of cuff links . Let x = original cost then from "marked up \$225 from cost, which amounts to a 50% increase" we get: .50x = 225 x = 225/.50 x = \$450 (original cost)
 Volume/193682: This question is from textbook geometry a sphere with radiur is inscribed in a cylinder. find the volume of the cylinder in terms of the radius?1 solutions Answer 145370 by nerdybill(7090)   on 2009-04-29 06:46:40 (Show Source): You can put this solution on YOUR website! a sphere with radiur is inscribed in a cylinder. find the volume of the cylinder in terms of the radius? . Radius of the cylinder is r Height of the cylinder is 2r . Volume of cylinder: (pi)r^2*h where pi is 3.14 r is radius h is height . Plugging in our values: (pi)r^2*h (3.14)r^2*2r 6.28r^3 (volume of cylinder)
 Linear-equations/193672: The linear equation: y=0.15x+0.79 What is the slope (or rate of change) of this equation?1 solutions Answer 145369 by nerdybill(7090)   on 2009-04-29 06:42:11 (Show Source): You can put this solution on YOUR website!The linear equation: y=0.15x+0.79 . The given equation is already in the "slope-intercept" form of: y = mx + b where m is the slope b is the y-intercept at (0,b) . So, slope is: 0.15 Or, if you like it as a fraction: 0.15 = 15/100 = 3/20
 logarithm/193652: log4(x)+log4(x-6)=21 solutions Answer 145355 by nerdybill(7090)   on 2009-04-28 22:18:43 (Show Source): You can put this solution on YOUR website! log4(x)+log4(x-6)=2 log4((x)(x-6))=2 (x)(x-6)=4^2 (x)(x-6)=16 x^2-6x=16 x^2-6x-16=0 (x-8)(x+2) = 0 . x = {-2, 8} Throw out the negative solution (can't have log of negatives): x = 8
 logarithm/193620: log base 4 (x-4) + log base 4 (x-2) = log base 4 351 solutions Answer 145334 by nerdybill(7090)   on 2009-04-28 20:27:02 (Show Source): You can put this solution on YOUR website!Apply log rules: log base 4 (x-4) + log base 4 (x-2) = log base 4 35 log base 4 (x-4)(x-2) = log base 4 35 (x-4)(x-2) = 35 x^2-2x-4x+8 = 35 x^2-6x+8 = 35 x^2-6x-27 = 0 (x-9)(x+3) = 0 . x = {-3, 9) We can toss out the negative solution because we can't take the ln of a negative number. So we are left with: x = 9
Quadratic_Equations/193575: I'm sorry. I know this problem is very similar to the example, but I get a very crazy answer when I've tried to do this like the example many times. I must be doing something wrong. Please help.
One leg of a right triangle is 96 inches. Find the hypotenuse and teh other leg if the length of the hypotenuse exceeds 2 and 1/2 times teh other leg by 4 inches.
1 solutions

Answer 145298 by nerdybill(7090)   on 2009-04-28 17:12:30 (Show Source):
You can put this solution on YOUR website!
One leg of a right triangle is 96 inches. Find the hypotenuse and teh other leg if the length of the hypotenuse exceeds 2 and 1/2 times teh other leg by 4 inches.
.
Let x = length of "other leg" of triangle
then
(5/2)x-4 = hypotenuse
.
The applying the Pythagorean theorem:
x^2 + ((5/2)x-4) = 96^2
x^2 + ((5/2)x-4)((5/2)x-4) = 96^2
x^2 + (25/4)x^2-20x +16 = 96^2
(29/4)x^2-20x +16 = 9216
(29/4)x^2-20x -9200 = 0
29x^2 - 80x -36800 = 0
Solving using the quadratic equation we get:
x ={37.03, -34.26}
Throwing away the negative solution we're left with:
x = 37.03 inches (other leg)
.
Hypotenuse:
(5/2)x-4 = (5/2)37.03-4 = 88.58 inches (hypotenuse)
.