What will be the equation of a circle concentric with the circle [math]x^2+y^2-4x-6y-9=0[/math] and passing through the point [math](-4,-5)[/math]?
Let us assume a equation of circle(x-a)^2+(y-b)^2=r^2Since the above circle is concentric with given circle i.e.,x^2+y^2-4x-6y-9=0The center of given circle will be (d,e)d=-(-4/2)e=-(-6/2)(d,e)=(2,3)The centre of assumed circle will also be (2,3) Because both are concentric circles .Therefore , we can say a=2,b=3 And circle is also passing through (-4,-5) Therefore(-4-2)^2+(-5-3)^2=r^2(-6)^2+(-8)^2=r^236+64=r^2r^2=100r=√100r=10Equation of circle will be (x-2)^2+(y-3)^2=100
Can you help with this differential equation?
No problem First rearrange the equations dy/dx= (64xy)^1/3 --> dy/dx= 4x^1/3*y^1/3 --> Then seperate by x's and y's dy(y^-1/3) = dx(4x^1/3) Now integrate both sides int(y^-1/3)dy = int(4x^1/3)dx --> 3/2y^(2/3) + c1 = 3x^(4/3) + c2 As per the response below you can simplify this even further. You begin by combining the c's by setting C = c2 - c1 So now you have 3/2y^(2/3) = 3x^(4/3) + C You can simplify this even more by solving for y y = (2x^(4/3) + (2/3)C)^(3/2) I hope this helps!
Dynamics - Differential Equations?
Newton's 2nd law of motion: m dv/dt = 32 m -- K v^2 ..........; m = 64 lb, v(0) = 10 ft/sec; K a constant The DE is separable: Int[ dv / ( 32 -- k v^2) ] from 10 to v(t) = t. Also k = K/64 The integral is standard. The limiting velocity determines K. The rest involves another integral.
In a test there are 80 questions. A candidate gets 1 mark for each correct answer, and a minus of .5 for each wrong answer. If the candidate scores 54.5 marks, how many correct answers did he have?
Getting one question wrong effectively costs the candidate 1.5 marks (the 1 mark he failed to earn and also the -.5 he earned for the wrong answer).54.5 marks out of 80 means that he failed to earn 25.5 marks. 25.5 / 1.5 gives the number of wrong answers.25.5/1.5 = 17→ 17 incorrect answers out of 80 questions means 63 were correct. ← answer
Which one performs better?AMD Athlon 64 X2 6000+or Intel core 2 duo E6600?
Don't say the intel one because ur an intel fanboy or amd cuz u'r an amd fan.I want to know generally, which one performs better( gaming,multitasking,web serfing,using office applications etc)?Btw athlonx2 6000+ costs 302$CAD and core 2 duo e6600 costs 373$CAD.Is the core 2 duo e6600 worth the extra 71$ over 6000+?or 6000+ is actually better and cheaper?Tnx.
What are the dimensions of a closed rectangular box that has a square cross section, a capacity of 135 in^3?
Let x be the length of the side of he squares. x^2 * y = 135 y = (135 / x^2) = (135*x^-2) <-------------- Critical later A = 2x^2 + (4xy) = 2x^2 + (540*x^-1) dA/dx = 4x - 540x^(-2) = 0 Multiply both sides by x^2, getting: 4x^3 - 540 = 0 4x^3 = 540 x^3 = 135, so x = 5.22193 <------------------- Answer Now we check the second derivative. Since the first derivative is dA/dx = 4x - 540x^(-2) d^2A/dx^2 = 4 + 1080/x^3...............which is postive for our answer. Thus we have minimized the area used. y = (135 / (5.22193)^2) = 4.95 The sides of the squares are 5.22 inches The third side is 4.95 inches .
Optimization problem, ?
the sides of a right triangle are on the coordinate axes and it's hypotenuse passes through the point (1,8) Find the vertices of this triangle such that the length of the hypotenuse is minimum
What would the algorithm for celsius to fahrenheit and vice versa be?
Well, C = (5/9)(F - 32) the reverse is easy... C : temp in Celsius F : temp in Fahrenheit i didn't knew it by heart before, and then i used the following "reasoning" to get it back in mind : the relation is linear and 0°C is 32°F and ..........95°F are 35°C so 95 - 32 = 63 = 7 * 9 ---> need to multiply by 5/9 to get 35°C ---> bizarre but it works ... ! ... it's the debate between units : as i ' m in aeronautics, we use terrestrial miles : 1604 meters (take-off speed...) knots : nautical miles : 1852m/hour flight level : FL390 = 39000 feet PSI : Pounds per Square Inch (Hydraulic pressures) etc. ;-) Bonne Année from France ! hope it' ll help !!