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Expressions-with-variables/174365: You have exactly $537 to spend on party gifts for your rich uncle’s birthday party. You decide to get watches for the ladies (at $27.98 each), and beepers for the men (at $23.46 each). You know that the number of watches required will be 3 times as much as the number of beepers. How many of each item do you buy? 1 solutions Answer 129307 by ankor@dixie-net.com(15633)   on 2008-12-22 19:59:35 (Show Source): You can put this solution on YOUR website! You have exactly $537 to spend on party gifts for your rich uncle’s birthday party. You decide to get watches for the ladies (at $27.98 each), and beepers for the men (at $23.46 each). You know that the number of watches required will be 3 times as much as the number of beepers. How many of each item do you buy? : Let x = no. of beepers then 3x = no. of watches : 23.46x + 27.98(3x) = 537 23.46x + 83.94x = 537 107.40x = 537 x = x = 5 beeper and 3*5 = 15 watches : : You can check solution in the original equation

Miscellaneous_Word_Problems/174316: At 12:00 noon Ship A is 80 miles due north of Ship B and is sailing north at a rate of 13 knots. Ship B is sailing east at a rate of 11 knots. Write the distance between the ships as a function of time, where T=0 represents 12:00 noon and determine how far apart (in miles) the two ships will be at 5pm (1 Knot = 1.15mph) 1 solutions Answer 129283 by ankor@dixie-net.com(15633)   on 2008-12-22 17:44:30 (Show Source): You can put this solution on YOUR website! At 12:00 noon Ship A is 80 miles due north of Ship B and is sailing north at a rate of 13 knots. Ship B is sailing east at a rate of 11 knots. Write the distance between the ships as a function of time, where T=0 represents 12:00 noon and determine how far apart (in miles) the two ships will be at 5pm (1 Knot = 1.15mph) : This is a Pythagorus problem: a^2 + b^2 = c^2 Where a = (5*13*1.15)+80 (travel time is 5 hrs for both ships) a = 154.75 mi and b = 5*11*1.15 b = 63.25 mi : c = distance apart in 5 hrs : c^2 = 154.75^2 + 63.25^2 c^2 = 23947.5625 + 4000.5625 c^2 = c = 167.177 mi apart in 5 hrs (5 pm) : the function of time equation: f(t) =

Surface-area/174119: Hello, I have a question that im extremely stuck on.The question is as follows:- A circular pool has a radius r. It is surrounded by a path which is 2 metres wide. Work out a formula for the area of the path and give its area when r=2.5 metres.Please could you show the step by step guide on how i would work this out.The more simplified the better. Many thanks 1 solutions Answer 129008 by ankor@dixie-net.com(15633)   on 2008-12-19 13:32:39 (Show Source): You can put this solution on YOUR website! A circular pool has a radius r. It is surrounded by a path which is 2 metres wide. Work out a formula for the area of the path: : The area (A1) of the pool: A1 = pi*r^2 ; The area (A2) of the pool and path together: The radius of this would be the pool radius plus 2 m; (r+2) : A2 = pi*(r+2)^2 : FOIL (r+2)^2 A2 = pi*(r^2 + 4r + 4) : Multiply everything inside the brackets by pi A2 = pir^2 + 4pir + 4pi (overall area) : The area of the path (Ap) is the overall area - the pool area: (Ap) = A2 - A1 so we have : Ap = (pir^2 + 4pir + 4pi) - pi*r^2 : Group like terms Ap = pir^2 - pir^2 + 4pir + 4pi; pir^2's cancel : Ap = 4pir + 4pi : Factor out 4pi: Ap = 4pi(r + 1); the formula ; : "and give its area when r = 2.5 metres" : Substitute 2.5 for r in the above equation Ap = 4pi(2.5+1) Ap = 4pi(3.5) Ap = 43.98 ~ 44 sq meters is the area of the path : : Check to see if this is true (overall radius; 2 + 2.5 = 4.5: Overall = pi*4.5^2 = 63.62 pool A = pi*2.5^2 = 19.63 ----------------------------- path area -----------44.00; confirms our solution ; Was this understandable to you? Any questions?

Surface-area/174099: Hello, Could you give me the surface area of a cylinder when the radius is 6cm and the Length is 21cm. Could you show me the full step by step guide to work this out. Many thanks 1 solutions Answer 129006 by ankor@dixie-net.com(15633)   on 2008-12-19 12:34:20 (Show Source): You can put this solution on YOUR website! Could you give me the surface area of a cylinder when the radius is 6cm and the Length is 21cm. : The surface area of cylinder consists of the parts Part 1: the height times the circumference (in this case it's the length) Part 2: the area of the two ends ; SA = (2*pi*r*h) + (2*pi*r^2) : In your problem; r=6, h=21 SA = (2*pi*6*21) + (2*pi*6^2) ; SA = 791.68 + 266.19 : SA = 1017.87 sq/cm

Travel_Word_Problems/174073: Doctor White leaves home to drive to a convention. She is traveling at 55miles/hour. She is 10 miles from home when her husband realizes that she has forgotten her briefcase. How fast will he have to drive to catch her in two hours? 1 solutions Answer 128985 by ankor@dixie-net.com(15633)   on 2008-12-19 07:59:56 (Show Source): You can put this solution on YOUR website! traveling at 55miles/hour. She is 10 miles from home when her husband realizes that she has forgotten her briefcase. How fast will he have to drive to catch her in two hours? : Find how far ahead White is at the start in terms of time (hrs): time = 10/55 time = 5/11 hrs : White travel time = 2 or hrs Husband's travel time = 2 hrs : Let s = speed required for him to catch up in two hrs ; When he catches up, they will have traveled the same distance Write a dist equation from this fact: (Dist = time * speed) : 2s = * 55 2s = 27(5) 2s = 135 s = 67.5 mph to catch W in 2 hrs ; : Check solution by find the distances: White's dist: 135 mi Husband's dist: 2*67.5 = 135 mi also

Quadratic_Equations/174047: The measure of the area of a rectangle is x2 -19x+48. If the dimensions of the rectangle are represented polynomials, find the demensions oftherectangle and then find its perimeter. 1 solutions Answer 128967 by ankor@dixie-net.com(15633)   on 2008-12-18 21:51:29 (Show Source): You can put this solution on YOUR website! The measure of the area of a rectangle is x^2 -19x+48. If the dimensions of the rectangle are represented polynomials, find the dimensions and then find it's perimeter. : Factor the equation to get the length and width x^2 - 19x + 48 = 0 (x-16)(x-3) = 0 x = 16, the length x = 3, the width : P = 2(16) + 2(3) P = 32 + 6 P = 38 units is the perimeter

Miscellaneous_Word_Problems/174033: A rectangle is three times as long as it is wide. If its lenght and width are both decreased by 2cm, its area is decreased by 36cm^2. Find its original dimensions. Make a sketch 1 solutions Answer 128940 by ankor@dixie-net.com(15633)   on 2008-12-18 19:50:30 (Show Source): You can put this solution on YOUR website! A rectangle is three times as long as it is wide. If its length and width are both decreased by 2cm, its area is decreased by 36cm^2. Find its original dimensions. Make a sketch : Let x = the width then 3x = the length and 3x^2 = original area : (x-2) = new width (3x-2) = new length and (x-2)*(3x-2) = new area FOIL (3x^2 - 8x + 4) = new area : Original area - new area = 36 sq/cm 3x^2 - (3x^2 - 8x + 4) = 36 3x^2 - 3x^2 + 8x - 4 = 36 8x = 36 + 4 x = x = 5 cm; original width then 3(5) = 15 cm; original length : : Check solution (new rectangle: 13 by 3)find the areas 75 - 39 = 36 : I'll let you do the sketching

Miscellaneous_Word_Problems/170376: Normal sea water is about 3.5% solids. This is about the same as saying "There are nearly 300 pounds of dissolved solid matter per thousand gallons of sea water" Using this information, find the weight of a gallon of sea water. THANK YOU SO MUCH! 1 solutions Answer 128936 by ankor@dixie-net.com(15633)   on 2008-12-18 19:34:15 (Show Source): You can put this solution on YOUR website! Normal sea water is about 3.5% solids. This is about the same as saying "There are nearly 300 pounds of dissolved solid matter per thousand gallons of sea water" Using this information, find the weight of a gallon of sea water. : Let x = weight of 1 gallon of sea water in pounds : .035 = x = 8.57 lb per gallon of sea water

Evaluation_Word_Problems/173958: 1. The perimeter of a rectangle is 120 feet. The width of the rectangle is equal to one half the length. Find the length and width of the rectangle? 2. The area of a rectangle is 576msqaured. The length of the rectangle is 4 times greater than the width. What are the length and width of the rectangle? 3, One integer is 8 less than another integer. The product of the lesser integer and -5 is 28 more than the greater integer. What are the integers? 1 solutions Answer 128881 by ankor@dixie-net.com(15633)   on 2008-12-18 16:33:38 (Show Source): You can put this solution on YOUR website! I'll give you the equations, see if you can do these now. : 1. The perimeter of a rectangle is 120 feet. The width of the rectangle is equal to one half the length. Find the length and width of the rectangle? Let x = width 2x = Length Perimeter equation 2(2x) + 2x = 120 Find x : : 2. The area of a rectangle is 576msqaured. The length of the rectangle is 4 times greater than the width. What are the length and width of the rectangle? x = the width 4x = the length Area equation 4x * x = 576 4x^2 = 576 Solve for x : 3, One integer is 8 less than another integer. The product of the lesser integer and -5 is 28 more than the greater integer. What are the integers? x = the larger integer (x-8) = the lesser integer Write an equation for the statement: "The product of the lesser integer and -5 is 28 more than the greater integer." -5(x-8) = x + 28 Find x : : If you have any questions about these, you can email me, but give it a try.

Expressions-with-variables/173966: This question is from textbook Algebra structure and method book 1 can you please help me solve: (((x+1/6 = 4/3))) 1 solutions Answer 128860 by ankor@dixie-net.com(15633)   on 2008-12-18 15:19:15 (Show Source): You can put this solution on YOUR website! can you please help me solve: : Multiply equation by 6 to get rid of the denominators; results: 6x + 1 = 2(4) : 6x = 8 - 1 x = ; : Check solution in original equation

Linear-equations/173902: How would I find the slope and y-intercept of this f(x)=-4x-8 1 solutions Answer 128780 by ankor@dixie-net.com(15633)   on 2008-12-17 19:45:48 (Show Source): You can put this solution on YOUR website! How would I find the slope and y-intercept of this f(x)=-4x-8 Very simple, the slope intercept form: y = mx + b where m = slope and b = y intercept : y = f(x) therefore in your equation: y = -4x - 8 therefore slope (m) = -4 and intercept (b) = -8

Miscellaneous_Word_Problems/173887: A house has two rooms of equal area. One room is square and the other room is a rectangle 4 ft narrower and 5 ft longer than the square room. Find the area of each room. 1 solutions Answer 128778 by ankor@dixie-net.com(15633)   on 2008-12-17 19:32:31 (Show Source): You can put this solution on YOUR website! : A house has two rooms of equal area. One room is square and the other room is a rectangle 4 ft narrower and 5 ft longer than the square room. Find the area of each room. : Let x = side of the square room then x^2 = area of the square room : (x+5) = length of the rectangular room (x-4) = width of the rectangular room Find the area of the rectangular room A = (x+5)*(x-4) = area FOIL A = x^2 + x - 20 : Rect area = square area x^2 + x - 20 = x^2 : x^2 - x^2 + x = +20 : x = 20 ft side of the square (A = 400 sq/ft) and 20+5 = 25 ft length of rect room 20-4 = 16 ft width of rect room (A = 400 sq/ft)

Geometry_Word_Problems/173891: The lengths of the sides of a triangle are consecutive even integers. Find the longest side is it is 22 units shorter than the perimeter. 1 solutions Answer 128776 by ankor@dixie-net.com(15633)   on 2008-12-17 19:20:24 (Show Source): You can put this solution on YOUR website! The lengths of the sides of a triangle are consecutive even integers. Find the longest side if it is 22 units shorter than the perimeter. : Let the three sides be: x, (x+2), (x+4) then the perimeter: p = x + (x+2) + (x+4) p = 3x + 6 : "Find the longest side is it is 22 units shorter than the perimeter." (3x+6 - (x+4) = 22 3x + 6 - x - 4 = 22 2x + 2 = 22 2x = 22 - 2 2x = 20 x = 10, the shortest side then 10 + 4 = 14 units is the longest side : : Check solution Perimeter - longest side = (10+12+14) - 14 = 22

Miscellaneous_Word_Problems/173876: A small city park consists of a rectangular lawn surrounded on all sides by 330 meters squared border of flowers 2.5 m wide. Find the area od the lawn if the entire park is 5 m longer than it is wide. 1 solutions Answer 128773 by ankor@dixie-net.com(15633)   on 2008-12-17 19:03:46 (Show Source): You can put this solution on YOUR website! A small city park consists of a rectangular lawn surrounded on all sides by 330 meters squared border of flowers 2.5 m wide. Find the area of the lawn, if the entire park is 5 m longer than it is wide. : Let x = width of the entire park then (x+5) = Length of the entire park therefore x(x+5) = area of the entire park : Since the walkway is 2.5 m wide, we can say: (x-5) = width of the lawn and (x+5)-5 = x; is the length of the lawn therefore x(x-5) = area of the lawn : We can say: entire park area - lawn area = 330 (area of the border) : x(x+5) - x(x-5) = 330 : x^2 + 5x - x^2 + 5x = 330 : x^2 - x^2 + 5x + 5x = 330 : 10x = 330 ; x = 33 m is the width of the entire area and 33 + 5 = 38 length of the entire area : Find the dimensions of the lawn 33 - 5 = 28 m width of the lawn and 33 m = length of the lawn : : Check solution by find the area of the entire park and the lawn (38*33) - (33*28) = 1254 - 924 = 330

Polynomials-and-rational-expressions/173857: This question is from textbook algebra book 1 mcdoul A rectangle is three times as long as it is wide. if its length and width are both decreased by 2cm, its area is decreased by 36cm squared(or raise to the 2nd power.) Find its original dimensions. 1 solutions Answer 128740 by ankor@dixie-net.com(15633)   on 2008-12-17 16:07:57 (Show Source): You can put this solution on YOUR website! A rectangle is three times as long as it is wide. if its length and width are both decreased by 2cm, its area is decreased by 36cm squared(or raise to the 2nd power.) Find its original dimensions. : Original rectangle: Let x = the width then 3x = the length and Area = 3x^2 : "if its length and width are both decreased by 2cm, its area is decreased by 36 sq/cm" (3x-2)*(x-2) = 3x^2 - 36 FOIL 3x^2 - 6x - 2x + 4 = 3x^2 - 36 3x^2 - 8x + 4 = 3x^2 - 36 3x^2 - 3x^2 - 8x = - 36 - 4 -8x = -40 x = x = +5 cm is the original length then 3*5 = 15 cm is the original length ; : Check solution in the statement: "its length and width are both decreased by 2cm, its area is decreased by 36cm" 13 * 3 = (15*5) - 36 39 = 75 - 36

Linear-equations/173784: Your cell phone plan charges you $.02 to send a text message and $.07 to recieve a text message. You plan to spend no more than $5 a month text messaging. a. Write an equation in standard form that models the possible combinations of sent text messages and received text messages. b. Graph the equation from part (a). Explain what the intercepts of the graph mean in this situation. c. List three other possible combinations of the number of messages you can send and receive. 1 solutions Answer 128712 by ankor@dixie-net.com(15633)   on 2008-12-17 13:42:28 (Show Source): You can put this solution on YOUR website! Your cell phone plan charges you $.02 to send a text message and $.07 to receive a text message. You plan to spend no more than $5 a month text messaging. : a. Write an equation in standard form that models the possible combinations of sent text messages and received text messages. : Let x = messages received let y = messages sent : .07x + .02y = 5 : b. Graph the equation from part (a). Explain what the intercepts of the graph mean in this situation. : convert equation to the general form .02y = 5 - .07x y = - x y = 250 - 3.5x : Plot these two points x | y ------- 20 |180 60 | 40 Your graph should look like this: Note: y axis are the no. messages sent, x axis are the no. of messages received ; The y intercept (250) means only sent messages, none received (x=0) The x intercept, approx 71 messages received, none sent (y=0) : : c. List three other possible combinations of the number of messages you can send and receive. you can do this; You can get the approximate x/y values from the graph or; To be exact, choose a value for x and find y in the above equation : Did this make sense to you. Any questions?

Linear-systems/173706: Please show me how to solve. I'm having trouble understanding from when I asked a few days ago. Solve by graphing. 3x + 4y ≤ 12 x + 3y ≤ 6 x ≥ 0 y ≥ 0 1 solutions Answer 128625 by ankor@dixie-net.com(15633)   on 2008-12-16 21:54:02 (Show Source): You can put this solution on YOUR website! Solve by graphing. 3x + 4y ≤ 12 x + 3y ≤ 6 x ≥ 0 y ≥ 0 : Plot the 1st two equations, put the equations in the slope/intercept form 3x + 4y =< 12 4y = 12 - 3x y = - x y = x + 3 : Calculate two points (substitute the values for x and find y): x | y ------- 0 | 4 4 | 1 Graph should look like this: : Do the same with the 2nd equation x + 3y => 6 3y = 6 - x y = - y = -x + 2 Calculate two points (substitute the values for x and find y): x | y ------- 0 | 2 3 | 1 Graph should look like this: : The last two equations tell you that only positive values for x and y are considered : Put both graphs on the same coordinate system: Look at this graph The 1st equation:3x + 4y =< ; 12, the purple line Area of feasibility is at or below this line : The 2nd equation: x + 3y => 6, green line Area of feasibility is at or above this line : x => 0 y => 0 As I said before, this means only positive values for x & y, inside the area of feasibility are considered. : The solution is the triangular area between the lines and to the right of the y axis.

logarithm/173741: Hello, My question concerns growth, but I do not understand how to use the log in the equation. It reads f(x)=400 log10 (2t+3). It is asking how many animals will there be after six months. Can you help me make sense of this problem? Thank you ever so much. 1 solutions Answer 128617 by ankor@dixie-net.com(15633)   on 2008-12-16 21:21:51 (Show Source): You can put this solution on YOUR website! My question concerns growth, but I do not understand how to use the log in the equation. It reads f(x)=400 log10 (2t+3). It is asking how many animals will there be after six months. : Assume you mean log,base10, which can just be written log( ), so we have f(t) = 400*log(2t+3) : Find f(t) when t = 6 f(t) = 400 * log[2(6) + 3] ; f(t) = 400 * log(12 + 3) : f(t) = 400 * log(15) : Find the log of 15 f(t) = 400 * 1.1761 : f(t) = 470 animals after 6 months : Hope this is what you mean.

expressions/173478: 1/(2X-4)- 2/(y+1)=0 and 1/(x-3)- 1/(y+4)=0 Solve using elimination: 1 solutions Answer 128614 by ankor@dixie-net.com(15633)   on 2008-12-16 21:08:32 (Show Source): You can put this solution on YOUR website! 1/(2X-4)- 2/(y+1)=0 and 1/(x-3)- 1/(y+4)=0 Solve using elimination: : - = 0 Multiply equation by (2x-4)(y+1) to get rid of the denominators, results (y+1) - 2(2x-4) = 0 y + 1 - 4x + 8 = 0 -4x + y + 9 = 0 -4x + y = -9 4x - y = 9; multiplied by -1 and - = 0 Multiply by (x-3)(y+4), results: (y + 4) - (x - 3) = 0 y + 4 - x + 3 = 0 -x + y + 7 = 0 -x + y = -7 : Add the two equations 4x - y = 9 -x + y = -7; ---------------adding eliminates y, find x 3x = 2 x = : Use the last equation to find y: - + y = -7 y = -7 + y = -6 or : : Check solutions in the 2nd original equation: - = 0 - = 0 - = 0 + = 0 : I'll let you check it in the 1st original equation

logarithm/173595: Log(base6)x+log(base6)(x-5)=2 Check for inadmissible roots 1 solutions Answer 128523 by ankor@dixie-net.com(15633)   on 2008-12-16 13:46:40 (Show Source): You can put this solution on YOUR website! Log(base6)x+log(base6)(x-5)=2 : log6(x) + log6(x-5) = 2 : adding logs mean multiply, so we can write it: log6(x(x-5)) = 2 : Write it in the exponential form x(x-5) = 6^2 : x^2 - 5x = 36 A quadratic equation x^2 - 5x - 36 = 0 Factor (x-9)(x+4) = 0 two solutions x = -4, not an admissible solution (can't have a log of a neg number) and x = +9, a good solution

Quadratic_Equations/173622: This question is from textbook Introductory Algebra Solve the formula for the indicated letter. S = (1/3nr^2h + 4nrh), for r 1 solutions Answer 128515 by ankor@dixie-net.com(15633)   on 2008-12-16 13:03:55 (Show Source): You can put this solution on YOUR website! Solve the formula for the indicated letter. S = (1/3nr^2h + 4nrh), for r : nr^2h + 4nrh = S Multiply equation by 3 to get rid of the denominator nr^2h + 3(4nrh) = 3S : nr^2h + 12nrh = 3S : Factor out nh: nh(r^2 + 12r) = 3S : r^2 + 12r = : r^2 + 12r - = 0 A quadratic equation where: a=1; b=12; c=- Use the quadratic formula (we only want the positive solution here:

Quadratic_Equations/173612: This question is from textbook Introductory Algebra The perimeter of a rectangle is 62 m. If the width were doubled and the length were increased by 21 m, the perimeter would be 120 m. What are the length and width of the rectangle? 1 solutions Answer 128496 by ankor@dixie-net.com(15633)   on 2008-12-16 10:25:15 (Show Source): You can put this solution on YOUR website! The perimeter of a rectangle is 62 m. If the width were doubled and the length were increased by 21 m, the perimeter would be 120 m. What are the length and width of the rectangle? : Write an equation for each rectangle, and use elimination to solve this: 2L + 2W = 62 Simplify, divide by 2: L + W = 31 : and "If the width were doubled and the length were increased by 21 m," P = 120 : 2(L+21) + 2(2W) = 120 2L + 42 + 4W = 120 2L + 4W = 120 - 42 2L + 4W = 78 Simplify, divide by 2 L + 2W = 39 : Subtract the 1st equation from the 2nd equation L + 2W = 39 L + W = 31 ----------- W = 8 m : Find L using L + W = 31 L = 31 - 8 L = 23 m : : Check solution by finding the perimeter of each 2(23) + 2(8) = 62 and 2(44) + 2(16) = 120

Expressions-with-variables/173608: I need to simpify this equation. 4+2/x / x/4+1/8 my answer was 16/x this just dosn't seem right can you help? 1 solutions Answer 128494 by ankor@dixie-net.com(15633)   on 2008-12-16 10:07:01 (Show Source): You can put this solution on YOUR website! I need to simpify this equation. 4+2/x / x/4+1/8 my answer was 16/x this just doesn't seem right can you help? : But you are right!!

Equations/173601: In sunday's 15 mile paddleboard race, after 6 miles, Kongo saw that he was 1/2 a mile behind his friend Kimo, whose speed was 6mph. At what rate did Kongo paddle, if he caught his friend exactly at the finish line? 1 solutions Answer 128492 by ankor@dixie-net.com(15633)   on 2008-12-16 10:00:37 (Show Source): You can put this solution on YOUR website! In Sunday's 15 mile paddle-board race, after 6 miles, Kongo saw that he was 1/2 a mile behind his friend Kimo, whose speed was 6mph. At what rate did Kongo paddle, if he caught his friend exactly at the finish line? : From the given information we know: Ko has to travel (15-6) 9 mi, while Ki has to travel (15-6.5) 8.5 mi : Let s = Ko's speed required to catch Ki at the finish line : Write a time equation; Time = (dist/speed) : Ko's time = Ki's time = Cross multiply 8.5s = 9 * 6 : 8.5s = 54 s = s = 6.353 mph Ko's speed required to accomplish this ; : Check solution by finding that the travel times are equal 9/6.353 = 1.41 hrs 8.5/6 = 1.41 hrs also

Percentage-and-ratio-word-problems/173517: if the length of a rectangle is increased by 10% and the area is unchanged , then by what percentage must the width be decreased ? 1 solutions Answer 128402 by ankor@dixie-net.com(15633)   on 2008-12-15 19:57:27 (Show Source): You can put this solution on YOUR website! if the length of a rectangle is increased by 10% and the area is unchanged , then by what percentage must the width be decreased ? : Let L & W be the original length and width : 1.1L = new length xW = new width : New area = old area 1.1L * xW = L * W Divide both sides by L 1.1 * xW = W Divide both sides by 1.1 xW = Divide both sides by W x = x ~ .91, original width reduced by 9%

Linear_Equations_And_Systems_Word_Problems/173499: Sam buys 3 kinds of fruit for 40 dollars, 10 dollars, and 1 dollar each. He pays 259 dollars for 100 pieces of fruit. How many of the cheapest did he buy? 1 solutions Answer 128392 by ankor@dixie-net.com(15633)   on 2008-12-15 19:18:55 (Show Source): You can put this solution on YOUR website! : Sam buys 3 kinds of fruit for 40 dollars, 10 dollars, and 1 dollar each. He pays 259 dollars for 100 pieces of fruit. How many of the cheapest did he buy? : Three kinds of fruit x, y, z : x + y + z = 100 and 40x + 10y + z = 259 : Subtract the 1st equation from the 2nd 40x + 10y + z = 259 x + y + z = 100 ---------------------- 39x + 9y = 159 Simplify, divide by 3 13x + 3y = 53 3y = 53-13x y = We know there can't be many $40 fruits (x) and y has to be an integer try x = 2 y = y = y = 9 ea $10 fruit, this is the only positive integer solution : Find z 2 + 9 + z = 100 z = 100 - 11 z = 89 ea $1 animals, (the cheapest fruit) : Check in the Cost equation 40(2) + 10(27) + 1(89) 80 + 90 + 89 = 259; confirms our solutions

Percentage-and-ratio-word-problems/173487: Mya played in two softball games one afternoon. The first game lasted 42 minutes. The second game lasted 1 2/3 times longer that the first game. How long did Mya's second game last? 1 solutions Answer 128349 by ankor@dixie-net.com(15633)   on 2008-12-15 16:51:04 (Show Source): You can put this solution on YOUR website! Mya played in two softball games one afternoon. The first game lasted 42 minutes. The second game lasted 1 2/3 times longer that the first game. How long did Mya's second game last? : Convert 1 2/3 to an improper fraction, namely ; * 42 = = 70 minutes

Miscellaneous_Word_Problems/173430: This question is from textbook introduction to college mathematics The decay rate of krypton-85 is 6.3% per day. What is the half-life? Using the exponential decay function P(t) =Po^e^-kt 1 solutions Answer 128347 by ankor@dixie-net.com(15633)   on 2008-12-15 16:44:28 (Show Source): You can put this solution on YOUR website! The decay rate of krypton-85 is 6.3% per day. What is the half-life? Using the exponential decay function P(t) =Po^e^-kt : The exponential decay formula that I am familiar with: A = Ao[2^(-t/h)] where Ao is the initial amt A = final amt t = time h = half-life of the substance : Assume Ao = 100, Assume t = 1 day, find h (half life in days) After 1 day, A = 100 - 6.3 = 93.7 : 100 * 2^(-1/h) = 93.7 : 2^(-1/h) = : 2^(-1/h) = .937 : ln[2^(-1/h)) = ln(.937) : *.693 = -.065 : = -.065 : h = h = 10.66 days is the half life of krypton-85 according to this information : In actual fact I think they mean 6.3% per year, hence, 10.66 yrs half life

Volume/173460: Jeremy has a fish tank that has 40-cm by 70-cm rectangular base. The water is 25-cm deep. When he drops rocks into the tank, the water goes up by 2-cm. What is the volume in liters of the rocks? 1 solutions Answer 128334 by ankor@dixie-net.com(15633)   on 2008-12-15 15:45:28 (Show Source): You can put this solution on YOUR website! Jeremy has a fish tank that has 40-cm by 70-cm rectangular base. The water is 25-cm deep. When he drops rocks into the tank, the water goes up by 2-cm. What is the volume in liters of the rocks? : Find the volume of water displaced in cu/cm : 40 * 70 * 2 = 5600 cu/cm ; Convert cu/cm to liters: = 5.6 liters

Volume/173211: A rectangular piece of metal 32cm by 22cm, has a square of side X cm removed from each corner in order to form a rectangular box. If the volume of the box is to be a maximum what is the value of X? 1 solutions Answer 128222 by ankor@dixie-net.com(15633)   on 2008-12-14 21:57:55 (Show Source): You can put this solution on YOUR website! A rectangular piece of metal 32cm by 22cm, has a square of side X cm removed from each corner in order to form a rectangular box. If the volume of the box is to be a maximum what is the value of X? : From the given information we know the dimensions (L,W,H) of the box is: (32-2x) by (22-2x) by x : Area = length * width * height A = (32-2x) * (22 - 2x) * x FOIL A = x(704 - 64x - 44x + 4x^2) A = x(704 - 108x + 4x^2) A = 704x - 108x^2 + 4x^3 or the standard arrangement is: y = 4x^3 - 108x^2 + 704x : Plot this equation, we only are interested in the positive values x,y values : Using my trusty Ti83, max volume occurs when x = 4.274 cm about 1348 cu/cm

Geometry_Word_Problems/173274: The lenght of a rectangle is 4cm more than its width and 4cm less than the length of a diagonal of the rectangle. What is the area of the rectangle? 1 solutions Answer 128203 by ankor@dixie-net.com(15633)   on 2008-12-14 20:46:51 (Show Source): You can put this solution on YOUR website! The length of a rectangle is 4cm more than its width and 4cm less than the length of a diagonal of the rectangle. What is the area of the rectangle? ; Keep in mind the right triangle formula a^2 + b^2 = c^2 : Let x = the length of the rectangle (a) then (x-4) = width of the rectangle (b) and (x+4) = diagonal of the rectangle (c) : x^2 + (x-4)^2 = (x+4)^2 FOIL x^2 + x^2 - 8x + 16 = x^2 + 8x + 16 Combine like terms on the left x^2 + x^2 - x^2 - 8x - 8x + 16 - 16 = 0 Results x^2 - 16x = 0 Factor x(x - 16) = 0 Two solutions; x = 0 and x = 16 is the length of the rectangle then 12 = width of the rectangle and 20 = diagonal of the rectangle ; Area = 16 * 12 = 192 sq/cm : Check it: 16^2 + 12^2 = 20^2 256 + 144 = 400; confirms our solution for length & width