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What else can I help you with?
the ratio of A to B is 2:3 and the ratio of B to C is 4:5. what is the ratio of A to C?
Oh, what a happy little problem we have here! To find the ratio of A to C, we can simply multiply the two ratios together. So, 2:3 times 4:5 gives us 8:15. That's the beautiful ratio of A to C, just like painting a lovely landscape with different colors blending harmoniously together.
What is 4 C B?
What is 4 C B
What is the answer to this ab - 2B plus ac -2C?
0
Which fraction is equivalent to 2 5. A 1 10 B 5 2 C 1 4 D 4 10?
B
What is the GPA for 2 A's 3 B's and 2 C's?
On a scale for GPA where A =4,B=3,C=2, and D=1 then GPA = (4 x2) + (3 x 3) + (2x2) all divided by 7 to get average GPA = 3.0
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)
Associative addition math problem?
a+(b+c)=b+(a+c) 2+(7+4)=7+(2+4)
If A is less than B and B plus C equals 10 and none of them equal zero then which of the following must be true?
You haven't provided any choices for the "which of the following" part of your question. Such questions are best avoided here. However, assuming a, b and c are all natural numbers, all of the following are true for a<b AND b+c=10: a=1, b=2, c=8 a=1, b=3, c=7 a=1, b=4, c=6 a=1, b=5, c=5 a=1, b=6, c=4 a=1, b=7, c=3 a=1, b=8, c=2 a=1, b=9, c=1 a=2, b=3, c=7 a=2, b=4, c=6 a=2, b=5, c=5 a=2, b=6, c=4 a=2, b=7, c=3 a=2, b=8, c=2 a=2, b=9, c=1 a=3, b=4, c=6 a=3, b=5, c=5 a=3, b=6, c=4 a=3, b=7, c=3 a=3, b=8, c=2 a=3, b=9, c=1 a=4, b=5, c=5 a=4, b=6, c=4 a=4, b=7, c=3 a=4, b=8, c=2 a=4, b=9, c=1 a=5, b=6, c=4 a=5, b=7, c=3 a=5, b=8, c=2 a=5, b=9, c=1 a=6, b=7, c=3 a=6, b=8, c=2 a=6, b=9, c=1 a=7, b=8, c=2 a=7, b=9, c=1 a=8, b=9, c=1
What is the GPA for 2 b's 4 c's and 1 d?
2.83 based on A=4, B=3, C=2, D=1. If each class has the exactly the same number of credits, then this would be a B+.
Are there any online tools for converting math powers Eg something that can tell me 2 to the power of 4 equals 4 to the power of 2?
Your example is confusing -- while 2^4 = 4^2, in general it is false that x^y=y^x, so not sure what you mean by "converting powers." A relevant rule is that (a^b)^c = a^(bc) (power of a power is product of powers) If a=2, b=2, and c=2 then a^b = 2-squared = 4 and the formula gives 4^2 = 2^(2*2) which is true. Also (a^b)*(a^c) = a^(b+c) (multiply numbers, add powers) and (a^b)/(a^c) = a^(b-c) (divide numbers, subtract powers)
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.
The average score on 4 tests is 85.5 the average of the three highest scores is 87 and her two lowest scores are the same number. what is the average of her two highest scores?
Let's take our four tests and call them A, B, C and D, in order from worst to best. We're told: (A + B + C + D) / 4 = 85.5 (B + C + D) / 3 = 87 A = B We can say then: (2B + C + D) / 4 = 85.5 Or: B/2 + C/4 + D/4 = 85.5 And: B / 3 + C / 3 + D / 3 = 87 Or: B/2 + C/2 + D/2 = 3(87 / 2) Now we can combine those two equations to eliminate B. We'll subtract the terms of the first one from the terms of the second one: (B/2 - B/2) + (C/2 - C/4) + (D/2 - D/4) = 3(87 / 2) - 85.5 (2C - C) / 4 + (2D - D) / 4 = 45 (C + D) / 4 = 45 (C + D) / 2 = 90 So the average of the two best tests is 90%
The first number of three consecutive even integers equals the sum of the second and third Find the three numbers?
Numbers are a, b and c; b = a + 2, c = a + 4. Find a such that a = b + c, ie a = (a + 2) + (a + 4) so a = 2a + 6 making a = -6, b = -4 and c = -2
the ratio of A to B is 2:3 and the ratio of B to C is 4:5. what is the ratio of A to C?
Oh, what a happy little problem we have here! To find the ratio of A to C, we can simply multiply the two ratios together. So, 2:3 times 4:5 gives us 8:15. That's the beautiful ratio of A to C, just like painting a lovely landscape with different colors blending harmoniously together.
Ax2 plus bx plus c equals 0?
x=(-b+sqrt(b^2-4*a*c))/(2a) and (-b-sqrt(b^2-4*a*c))/(2a) (This is called the quadratic formula)
How do you write an equation for a parabola with a vertex at -4 2 and y-intercept -2?
1. The y-intercept of a parabola with equation y = ax^2 + bx + c is c. So, c = -22. The vertex is (x, y) = (-4, 2), where x = - b/2ax = - b/2a-4 = - b/2a(-4)(-2a) = (-b/2a)(-2a)8a = bSo we have:y = ax^2 + bx + c (substitute what you know: 2 for y, -4 for x, 8a for b, and -2 for c)2 = a(-4)^2 + (8a)(-4) + (-2)2 = 16a - 32a - 22 = - 16a - 2 (add 2 to both sides)4 = -16a (divide by -16 to both sides)-1/4 = aSince b = 8a, then b = 8(-1/4) = -2 Since a = -1/4, b = -2, and c = -2, then we can write the equation of the parabola asy = (-1/4)x^2 - 2x - 2.
What is the answer to Find the coordinates of the reflection B (4 2) is reflected over the y-axis B and rsquo Question 9 options a) (4 -2) b) (-4 2) c) (-4 -2) d) (4 2) Save?
The answer is b.
