Tutors Answer Your Questions about Rectangles (FREE)
Question 1183445: Rectangle EFGH's base is b, its height is h, and it has a perimeter of 48 units. Determine the area of the rectangle for b=6, b=8, b=12, and b=16. Based on these results, what conjecture can you make about maximizing the area of a rectangle with a fixed perimeter?
Click here to see answer by greenestamps(13200)   |
Question 1185880: An object is thrown upward from a height of 80 ft. The initial velocity of the object is 64 ft per second. If the height h (in feet) is h=-16 +64t+80, where t is time in seconds, when will the object reach the ground?
Click here to see answer by Alan3354(69443)  |
Question 1185880: An object is thrown upward from a height of 80 ft. The initial velocity of the object is 64 ft per second. If the height h (in feet) is h=-16 +64t+80, where t is time in seconds, when will the object reach the ground?
Click here to see answer by ikleyn(52778)  |
Question 1186201: Raine is grass cutting a rectangular field which has a length of 40 meters and a width of 30 meters. She left an uncut area with the
dimensions:length-20 m and width-15 m.How much is the fractional part of the lawn that remains uncut?
Click here to see answer by ikleyn(52778) |
Question 1187268: NASA launches a rocket at t=0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t)= −4.9t2 squared + 325t + 353.
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur? For intermediate work, keep at least five decimal places. For the final answer, round to the nearest hundredth.
The rocket splashes down after seconds.
How high above sea-level does the rocket get at its peak?
The rocket peaks at meters above sea-level.
Click here to see answer by Alan3354(69443) |
Question 1187268: NASA launches a rocket at t=0 seconds. Its height, in meters above sea-level, as a function of time is given by h(t)= −4.9t2 squared + 325t + 353.
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur? For intermediate work, keep at least five decimal places. For the final answer, round to the nearest hundredth.
The rocket splashes down after seconds.
How high above sea-level does the rocket get at its peak?
The rocket peaks at meters above sea-level.
Click here to see answer by ikleyn(52778) |
Question 1187267: An object is thrown upward at a speed of 174 feet per second by a machine from a height of 5 feet off the ground. The height of the object after t seconds can be found using the equation h= −16t2 squared + 174t + 5.
When will the height be 119 feet?
When will the object reach the ground?
Click here to see answer by Alan3354(69443) |
Question 1187267: An object is thrown upward at a speed of 174 feet per second by a machine from a height of 5 feet off the ground. The height of the object after t seconds can be found using the equation h= −16t2 squared + 174t + 5.
When will the height be 119 feet?
When will the object reach the ground?
Click here to see answer by ikleyn(52778) |
Question 1187270: A certain electronics manufacturer found that the average cost C to produce x DVD/Blu- ray players can be found using the equation C=0.04x2 squared −4x + 800
. What is the minimum average cost per machine and how many DVD/Blu-ray players should be built in order to achieve that minimum?
The number of DVD/Blu-ray players that should be build to achieve the minimum is
The minimum average cost is
**Answer both to the nearest whole number.
Click here to see answer by ankor@dixie-net.com(22740)  |
Question 1187561: Can someone please help me with this problem? I’ve been struggling with it for a while now and my Geometry textbook doesn’t explain it very well.
Here is the problem:
Brett is buying a rug for his living room. His living room is a rectangle with a length that is twice as long as its width. Brett decides that the rug will also be a rectangle, but it will be only one half the length of the room and one half as wide. What will be the ratio of the rug’s area to the living room’s area?
Click here to see answer by greenestamps(13200) |
Question 1187561: Can someone please help me with this problem? I’ve been struggling with it for a while now and my Geometry textbook doesn’t explain it very well.
Here is the problem:
Brett is buying a rug for his living room. His living room is a rectangle with a length that is twice as long as its width. Brett decides that the rug will also be a rectangle, but it will be only one half the length of the room and one half as wide. What will be the ratio of the rug’s area to the living room’s area?
Click here to see answer by Theo(13342)  |
Question 1187110: Brett is buying a rug for his living room. His living room is a rectangle with a length that is twice as long as its width. Brett decides that the rug will also be a rectangle, but it will be only one half the length of the room and one half as wide. What will be the ratio of the rug’s area to the living room’s area?
Click here to see answer by Boreal(15235)  |
Question 1189165: Dear Colleagues
I have the following question to find a solution to:
The length of a rectangle exceeds its breadth by 4 centimetres. If the length were halved and the breadth increased by 5 centimetres, the area would be decreased by 35 square centimetres. Find the length of the rectangle.
I do not know if I can draw in this website, but whilst trying to find the solutions, I drew 2 rectangles: the one on the left had the dimensions (x)(x+4) for width and length respectively, the one on the right had dimensions ((x+4)/2)(x+5) for length and width respectively.
The question then reads - for me - [((x+4)/2)(x+5)] - 35 = x(x+4)
Multiplying the left hand by 2 to clear the fraction, we get
[(x+4)(2x+10)]-35 = x(x+4)
and after multiplying out we get
[2x^2+10x+8x+40]-35 = x^2+4x
Collecting like terms gives us
2x^2+18x+5 = x^2+4x
which simplifies to
x^2+14x+5=0
We cannot factorise, so using the quadratic formula we get
a = 1, b= 14, c = 5
and putting all of this into the formula we get
x = (-b+/- sqrt of b^2-4ac)/2
which gives
x = (-14 +/- sqrt (14)^2-(4)(1)(5))/2
which gives us, finally, x = (-14 +/-13.23)/2
I am assuming that we need a real number for the length of the side of the rectangle, so do not wish to get into imaginary numbers...
Where have I gone wrong?
Click here to see answer by ikleyn(52778) |
Question 1189165: Dear Colleagues
I have the following question to find a solution to:
The length of a rectangle exceeds its breadth by 4 centimetres. If the length were halved and the breadth increased by 5 centimetres, the area would be decreased by 35 square centimetres. Find the length of the rectangle.
I do not know if I can draw in this website, but whilst trying to find the solutions, I drew 2 rectangles: the one on the left had the dimensions (x)(x+4) for width and length respectively, the one on the right had dimensions ((x+4)/2)(x+5) for length and width respectively.
The question then reads - for me - [((x+4)/2)(x+5)] - 35 = x(x+4)
Multiplying the left hand by 2 to clear the fraction, we get
[(x+4)(2x+10)]-35 = x(x+4)
and after multiplying out we get
[2x^2+10x+8x+40]-35 = x^2+4x
Collecting like terms gives us
2x^2+18x+5 = x^2+4x
which simplifies to
x^2+14x+5=0
We cannot factorise, so using the quadratic formula we get
a = 1, b= 14, c = 5
and putting all of this into the formula we get
x = (-b+/- sqrt of b^2-4ac)/2
which gives
x = (-14 +/- sqrt (14)^2-(4)(1)(5))/2
which gives us, finally, x = (-14 +/-13.23)/2
I am assuming that we need a real number for the length of the side of the rectangle, so do not wish to get into imaginary numbers...
Where have I gone wrong?
Click here to see answer by Theo(13342) |
Question 1189165: Dear Colleagues
I have the following question to find a solution to:
The length of a rectangle exceeds its breadth by 4 centimetres. If the length were halved and the breadth increased by 5 centimetres, the area would be decreased by 35 square centimetres. Find the length of the rectangle.
I do not know if I can draw in this website, but whilst trying to find the solutions, I drew 2 rectangles: the one on the left had the dimensions (x)(x+4) for width and length respectively, the one on the right had dimensions ((x+4)/2)(x+5) for length and width respectively.
The question then reads - for me - [((x+4)/2)(x+5)] - 35 = x(x+4)
Multiplying the left hand by 2 to clear the fraction, we get
[(x+4)(2x+10)]-35 = x(x+4)
and after multiplying out we get
[2x^2+10x+8x+40]-35 = x^2+4x
Collecting like terms gives us
2x^2+18x+5 = x^2+4x
which simplifies to
x^2+14x+5=0
We cannot factorise, so using the quadratic formula we get
a = 1, b= 14, c = 5
and putting all of this into the formula we get
x = (-b+/- sqrt of b^2-4ac)/2
which gives
x = (-14 +/- sqrt (14)^2-(4)(1)(5))/2
which gives us, finally, x = (-14 +/-13.23)/2
I am assuming that we need a real number for the length of the side of the rectangle, so do not wish to get into imaginary numbers...
Where have I gone wrong?
Click here to see answer by greenestamps(13200) |
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