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What is the area in square millimeters of a right triangle with a hypotenuse of 68 millimeters and a leg of 60 millimeters?
The area is 960mm2To find the other angle you use pythagoras' theorem.682=602+x24624=3600+x2x2=1024x= 32The area of a triangle is half length by height:32/2 is 16. 16x60 = 960mm2
How do you solve x plus 9 equals x squared?
By factorising:x + 9 = x2x2 - x - 9 = 0(x - 3.54)(x + 2.54) = 0x = 3.54, x = -2.54By the quadratic formula:x = -(-1) ± √[(-1)2 - 4(1)(-9)]/2(1)x = 1 ± √(1 - -36)/2x = 1 ± √37/2x = 0.5 ± 6.08/2x = 0.5 + 3.04 = 3.54, x = 0.5 - 3.04 = -2.54The same solutions are given both ways, which confirms the result.
How do you make 52 using 1 2 3 4?
(4x3+1)x2x2=52
What is the prime factorization of 3x3x3 x2x2?
2 x 2 x 3 x 3 x 3 is the prime factorization of 108.
What is the area in square millimeters of a right triangle with a hypotenuse of 68 millimeters and a leg of 60 millimeters?
The area is 960mm2To find the other angle you use pythagoras' theorem.682=602+x24624=3600+x2x2=1024x= 32The area of a triangle is half length by height:32/2 is 16. 16x60 = 960mm2
What does power of 3 mean in math?
A number to the power of 3 (colloquially to the third power) refers to raising the number by an exponent of 3. An exponent of a number represents how many times that number is multiplied by itself.For example:2 to the power of 3 is the same as 23, which is the same as 2 x 2 x 2 = 8.
What is the highest common factor for 24 32 and 80?
hcf(24, 32, 80) = 8 24 = 2^3 x 3 32 = 2^5 80 = 2^4 x 5 hcf = 2^3 = 8
How do you solve x plus 9 equals x squared?
By factorising:x + 9 = x2x2 - x - 9 = 0(x - 3.54)(x + 2.54) = 0x = 3.54, x = -2.54By the quadratic formula:x = -(-1) ± √[(-1)2 - 4(1)(-9)]/2(1)x = 1 ± √(1 - -36)/2x = 1 ± √37/2x = 0.5 ± 6.08/2x = 0.5 + 3.04 = 3.54, x = 0.5 - 3.04 = -2.54The same solutions are given both ways, which confirms the result.
What are the main issues related to the numerical solution of two-point boundary value problems?
In the theory of ODEs (ordinary differential equations), an initial value problem (IVP)y'(t)=f(t,y), y(a)=c specifies a unique condition at the point a.A two-point boundary value problem (2PBVP) specifies conditions at two points (a and b):y''(t)=f(t,y,y'), y(a)=c y(b)=dAs usual, you can transform a second order ODE into a system of two first order ODEs, by defining:x1=yx2=y'so that:x1'(t)=x2x2'(t)=f(t,x1,x2)The problem is that, while one of the two conditions, say x1(a)=c, remains valid, you are not able to translate the other, say x1(b)=d into a condition on x2(b). Hence, what you do is to create a dummy condition on x2(b), say x2(b)=e, and then you numerically solve the system for different values of e, until you find a solution that also satisfies the condition x1(b)=d.
