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McDougal Littell Wordskills green level answers?
unit 11. b 2. d 3. c 4. d 5. a 6. c 7.d 8. b 9. d 10.a 11. d 12. a 13. c 14. b 15. d 16. d 17. b 18. d 19. a 20. d
Which best describes the form of a blues text?
A-B-C
What is the definition of the transitive property?
If a then b then b then c etc
What are the answers to the reading plus level f?
The answers can be found below. 1.a 2.c 3.a 4.d 5.d 6.b 7.c 8.a 9.d 10.c
Dashing through the snow?
The solfech is do la sol fa do, do do do la sol fa re. That is only the first part by syllables. "dashing through the snow in a one horse open sleigh." The notes really depend on what key you want to play or sing the song in. usually it is played in the key of C. so it would be C A G F C, C C C A G F D. that is the only part i can help with! sorry~ low D B A G D D D D B A G E E C B A F# high D D C A B low D B A G D D B A G E E C B A high D D D D E D C A G
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)
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.
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)
Evaluate a - b - c if a equals -15 b equals 5 and c equals -2?
If a = -15, b = 5 and c = -2 a - b - c = -15 - 5 - (-2) = -20 + 2 = -18
How do you find a leg in a right triangle?
In a right the triangle with legs a, b and hypotenuse c, a^2 = c^2 - b^2 or b^2 = c^2 - a^2.
What is the formula for a Venn Diagram?
A or B or C = A + B + C - A and B - A and C - B and C - 2 (A and B and C) I'm not sure by the way;
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.
Asquare plus b square plus c square -ab -bc -ca equals 0 then show that a equals b equals c?
a^2 + b^2 + c^2 - ab - bc - ca = 0=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0 => a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca + a^2 = 0 => (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 Each term on the left hand side is a square and so it is non-negative. Since their sum is zero, each term must be zero. Therefore: a - b = 0 => a = b b - c = 0 => b = c.
What is (a-b plus c)2?
The Pythagoream Thereom is a^2 + b^2 = c^2. Written out it is a squared plus b squared equals c squared.
What is the surface area of a cuboid?
The answer should be: (2*a*b)+(2*b*c)+(2*c*a)
