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What does c mean in c x 2 equals b?
b divided by 2
What does 2 c b t a c mean?
2 x b x t x a x c to the power 2
A minus b plus c whole square?
If you mean (a-b+c)^2, then... a^2 - ab + ac - ab + b^2 - bc + ac - bc + c^2 = a^2 + b^2 + c^2 - 2ab + 2ac - 2bc.
What does B plus A mean at bottom of a letter?
b=2 a=1 c=3 so b plus a =c
What does b mean in c x 2 equals b?
An unknown number, we'll call it B, is the same number as an unknown number C times 2.
What is proof of Heron's Formula?
This is a proof that uses the cosine rule and Pythagoras' theorem. As on any triangle with c being the opposite side of θ and a and b are the other sides: c^2=a^2+b^2-2abcosθ We can rearrange this for θ: θ=arccos[(a^2+b^2-c^2)/(2ab)] On a right-angle triangle cosθ=a/h. We can therefore construct a right-angle triangle with θ being one of the angles, the adjacent side being a^2+b^2-c^2 and the hypotenuse being 2ab. As the formula for the area of a triangle is also absinθ/2, when a and b being two sides and θ the angle between them, the opposite side of θ on the right-angle triangle we have constructed is 4A, with A being the area of the original triangle, as it is 2absinθ. Therefore, according to Pythagoras' theorem: (2ab)^2=(a^2+b^2-c^2)^2+(4A)^2 4a^2*b^2=(a^2+b^2-c^2)^2+16A^2 16A^2=4a^2*b^2-(a^2+b^2-c^2)^2 This is where it will start to get messy: 16A^2=4a^2*b^2-(a^2+b^2-c^2)(a^2+b^2-c^2) =4a^2*b^2-(a^4+a^2*b^2-a^2*c^2+a^2*b^2+b^4-b^2*c^2- a^2*c^2-b^2*c^2+c^4) =4a^2*b^2-(a^4+2a^2*b^2-2a^2*c^2+b^4-2b^2*c^2+c^4) =-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.1) We will now see: (a+b+c)(-a+b+c)(a-b+c)(a+b-c) =(-a^2+ab+ac-ab+b^2+bc-ac+bc+c^2)(a^2+ab-ac-ab-b^2+bc+ac+bc-c^2) =(-a^2+b^2+2bc+c^2)(a^2-b^2+2bc-c^2) =-a^4+a^2*b^2-2a^2*bc+a^2*c^2+a^2*b^2-b^4+2b^3*c-b^2*c^2+2a^2*bc-2b^3*c+(2bc)^2-2bc^3+a^2*c^2-b^2*c^2+2bc^3-c^4 =-a^4+2a^2*b^2+2a^2*c^2-b^4+(2bc)^2-c^4-2b^2*c^2 =-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.2) And now that we know that Eq.1=Eq.2, we can make Eq.1=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) Therefore: 16A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16 =[(a+b+c)/2][(-a+b+c)/2][(a-b+c)/2][(a+b-c)/2] And so if we let s=(a+b+c)/2 A^2=s(s-a)(s-b)(s-c)
Formula of a² plus b²?
If you mean: a^2 + b^2 = c^2 then it is Pythagoras' theorem or formula for a right angle triangle
What did the pythagorean theorm do for math?
If you know two sides of a right angle triangle, you can figure out the third by using the following formula: A*A+B*B=C*C or A**2+B**2=C**2 or C*C-B*B=A*A or C**2-B**2=A**2 or C*C-A*A=B*B or C**2-A**2=B**2 A is the bottom side if the right angle is at the bottom left, B is The only vertical(strait up and down) line if A is correct, and C is the last line.
Show that among all rectangles with area A the square has the minimum perimeter?
Suppose sqrt(A) = B ie the square with sides B has an area of A and its perimeter is 4*B. Now consider a rectangle with sides C and D whose area is A. So C*D = A = B*B so that D = B*B/C Perimeter of the rectangle = 2*(C+D) = 2*C + 2*D = 2*C +2*B*B/C Now consider (C-B)2 which, because it is a square, is always >= 0 ie C*C + B*B - 2*B*C >= 0 ie C*C + B*B >= 2*B*C Multiply both sides by 2/C (which is >0 so the inequality remains the same) 2*C + 2*B*B/C >= 4*B But, as shown above, the left hand side is perimeter of the rectangle, while the right hand side is the perimeter of the square.
What 3 numbers have a mean of 10 range of 2 and a median of 10.3?
So we have: the mean; (a + b + c)/3 = 10 yield that a + b + c = 30 (numbers in order) the range; c - a = 2 yield that c = a + 2 the median; b = 10.3 Substituting this information, we have: a + 10.3 + a + 2 = 30 2a = 17.7 a = 17.7/2 a = 8.85 yield that c = 10.85 So the numbers are 8,85, 10.3, and 10.85. Or, since b = 10.3 yield that a + c = 19.7 a + c = 19.7 -a + c = 2 add both equations 2c = 21.7 c = 18.85 which yield a = 8.85
Rearrange a equals 2b - c to make b the subject?
a = 2b - c a + c = 2b (a+c)/2 = b b = (a+c)/2
How do you complete a square in a math problem?
Here are the steps: ax^2 + bx + c = 0 Subtract c and divide by a x^2 + (b/a)x = -(c/a) Take the square of (b/a)/2 and add it to both sides (x + ((b/a)/2))^2 = -(c/a) + ((b/a)/2)^2 Take the square root of both sides Subtract ((b/a)/2) and you have your solutions: x = -(c/a) + ((b/a)/2)^2 - ((b/a)/2) x = (c/a) - ((b/a)/2)^2 - ((b/a)/2)
