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
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.
(4x3+1)x2x2=52
2 x 2 x 3 x 3 x 3 is the prime factorization of 108.
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
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.
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.
The highest common factor (HCF) of 24, 32, and 80 is 8. To find the HCF, you need to determine the common factors of each number and then identify the highest one they all share. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. The factors of 32 are 1, 2, 4, 8, 16, and 32. The factors of 80 are 1, 2, 4, 5, 8, 10, 16, 20, 40, and 80. The highest common factor among these numbers is 8.
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.