MATH QUESTION EXPERTS ONLY?
Adult appetite = 5/8 School age appetite = adult/ (1 3/4) = (5/8) / 7/4) = 5/8 * 4/7 = 5/14 (To check that, multiply 5/14 * 7/4 = 35/56 = 5/8.) Baby appetite = 1/2 (5/14) = 5/28 Now, just multiply by the number of each type and add: 56 (5/8) = 35 14 (5/14) = 5 28 (5/28) = 5 Total sandwiches needed: 45 That's it!
Math question for math experts?
Owner lost the $30 dollars for sure, as that was cash. But he also lost the product/goods worth $70 dollars. Talking strictly about retail price in profit/loss condition, I need to know the wholesale price of the product the lady bought. So cumulatively, the shop owner lost $30 dollars cash + the price HE bought the goods for (also known as the cost price)
Experts Help With Math Question !?
If it is (1/2)x - 6y = 10 then x - 12y = 20 => x = 20 + 12y so its not A 6y = (1/2)x + 10 => y = (1/12)x + 10/6 => y = (1/12) + 5/3 Answer: C Helpful note Use brackets to make expressions clear: 1/2x could be (1/2)x or 1/(2x).
Question for math Experts?
Aristotle Winger Dr. Aristotle Winger obtained his Ph. D on the age of 26 from Carnegie Mellon college interior the element of Mathematical Sciences. A graduate of the Howard college type of 1998, he's plenty extra suitable than purely a mathematician. He has been Professor of arithmetic at Emory and Henry college in Southwestern Virginia because of the fact the fall of 2004.
HELP MaTH QUESTION MATH EXPERTS HERE!?
1.) First you would need to find the total test scores. To do this you multiple the average test score by the total number of students. This is true because the meaning of average is the total number multiplied by the total set of numbers equals to your average. After you find the average of each set of test scores, you are going to minus the first set from the second set. With the answer you get, you divide the number by 4. Right now your answer is the combination of 4 tests, so dividing it by four would give you the average. 84 * 20 = 1680 <---1st set of test scores total 88 * 16 = 1408 <---2nd set of test scores total 1680-1408 = 272 <---Total difference in the two test scores 272/4 = 68 <---Divide by 4 since four tests were taken out 68 was the average scores eliminated 2.) 1cm = 100m p = meters 1 cm = 100p 100p 3.) Find all the factors of 80. The easiest way to do this is start with 1 and work your way up. When you write it out make sure to write it like (1x80) that way you know both 1 and 80 are factors of 80. factors of 80: 1x80 2x40 4x20 5x16 8x10 Add all the numbers up: 1+2+4+5+8+10+16+20+40+80 = 186 The sum of all the distinct factors of 80 is 186
Math question! Experts help please?
You need to set up a system of equations. One equation depicts the total cost of the mixture, and another depicts the total weight of the mixture. Let x= Pounds of high-quality Let y= Pounds of low-quality x+y=180 (since the total pounds must equal 180) 5.5x+3y= 808.20 808.20 comes from multiplying 4.49*180 (the total cost of the mixture) Using this system, solve for x and y (hopefully something you can do), and x will be how many pounds of high-quality to use, and y will be how many pounds of low-quality to use.
Question about Quadratics? ( math experts?)?
example: let y = 80x² - 2x - 63 you want to solve this type equation, right? ************************* to find the y - intercept, x = 0 y = 80(0)² - 2(0)² - 63 = -63 so y - intercept = -63 ****************************** to find the the x - intercept, y = 0 0 = 80x² - 2x - 63 ------------------------ method 1: factoring ------------------------ 0 = (10x - 9)(8x + 7) 10x - 9 = 0 or 8x + 7 = 0 10x = 9 or 8x = -7 x = 9/10 or -7/8 so the x - intercept is 9/10 and -7/8 ------------------------ method 2: completing the square ------------------------ 0 = 80x² - 2x - 63 80x² - 2x - 63= 0 divide the whole equation by 80 (coefficient of 80) x² - x/40 - 63/80 = 0 (x - 1/80)² - (1/80)² - 63/80 =0 simplify: (x -1/80)² - 5041/6400 = 0 (x - 1/80)² = 5041/6400 x - 1/80 = 71/ 80 or -71/80 x = 72/80 or -70/80 x = 9/10 or -7/8 --------------- method 3: use quadratic formula: --------------- a = 80, b = -2, c = -63 x = (-b ± √(b² - 4ac)) / 2a you would get x = 9/10 and -7/8 ************************** discriminant is b² - 4ac if the discriminant is: 1. negative, it means no real roots 2. zero, it means 1 root 3. positive, it means 2 distinct real roots ********************* solving vertex:::::: ---------------------------- method 1: completing square: ---------------------------- from above: (x -1/80)² - 5041/6400 so the vertex is at coordinate (1/80, -5041/6400) concaves down because coefficient of x² is positive ---------------------------- method 2: differentiation ---------------------------- y = 80x² - 2x - 63 dy/dx = 160x - 2 dy/dx = 0 160x - 2 = 0 x = 1/80 y = 80(1/80)² - 2(1/80) - 63 = -5041/6400 d²y/dx² = 160 {positive} so it concave up coordinate of vertexis (1/80, -5041/6400) **************************** axis of symmetry is x = -b/a b = 2, a = 80 so axis of symmetry is x = 2/80 x = 1/40 ************************* that's all
Which math expert on Quora is the best at math?
I agree with the other answerer that technically the answer is probably Cédric Villani, but he doesn’t really count since he wrote only one answer, a while ago, and isn’t active anymore.I think the active user with the most expertise is Qiaochu Yuan. He’s a grad student, so he’s actively doing research. Since he works on algebraic topology, he must be intimately familiar with abstract algebra, differential geometry, and category theory, while, with apologies to some of the other people listed, they might be brilliant but knowledge of these things fades over time. You can see his answers at MO here, and I doubt any of the other people listed would be able to answer a tenth of the questions he answers.But in terms of problem solving ability the winner is likely to be Anders Kaseorg, who is an IMO gold medallist and somehow manages to often find exceptionally elegant ways of solving problems posed here on Quora.
A few math questions for math experts? Please explain!?
1. Combine the like terms since everything is added you do not need parenthesis: 7x+10+ (-2x) +5 since 7x + (-2x) is the same as 7x-2x: 7x-2x +10 +5 5x+15 2. Multiplying everything: when the letters are already there, you add up the exponents. If you do not see an exponent, it is one. Also, negative x negative = positive, multiply the numbers (coefficients) together as well 6ab^2(-a)(-5a) = (6)(-1)(-5)(a)(a)(a)(b^2) = 30a^3b^2 3. (3x+4)(2x-2) use the distributive property, some people call it FOIL - multiply the First Terms, Outer ones, Inner ones and the Last ones; (3x)(2x) + (3x)(-2) + (4)(2x) + (4)(-2) =6x^2 + (-6x) + 8x + (-8) = 6x^2 -6x +8x -8 combine like terms; = 6x^2 +2x - 8 final answer 4. 9x^3+6x^2+3x / 3x well as written, the last term 3x/3x = 1 and that simplifies to 9x^3+6x^2+1, however if the 3x in the denominator (bottom) was under the entire 9x^3+6x^2+3x then it is this: (9x^3+6x^2+3x) / 3x (need the parenthesis because division is normall done before adding and subtracting) and the answer is to seperate each of the terms out; (9x^3)/3x + (6x^2)/3x +(3x) / 3x and when dividing, divide the numbers and when there is a variable, subtract the exponent (if there is no exponent, it is one) 3x^2 + 2x + 1 which is the answer you probably want 5. There are two terms there that cannot be combined and that makes it binomial. monomial is one term, tri = 3 terms poly is a number of terms 1, 2, 3, etc 6. (x+11y)-(-2x+4y) the - between them is like a -1 being there so distribute first: (x+11y) + (-1)(-2x+4y) (x+11y) + ( (-1)(-2x) + (-1)(4y) ) (x+11y) + (2x + (-4y) ) (x+11y) + (2x - 4y) and all are add or subtract so combine like terms x+11y + 2x - 4y x + 2x + 11y - 4y 3x + 7y final answer hope that helps
Do you have to be an expert in math for a BCA?
Not at all buddy :)I had the same question when i was thinking to join BCA. But as I am doing BCA now(currently pursuing 6th sem) I can say forget about the “Expert in math” thing, even if you are a average student in math then there will be no problem for you in pursuing BCA. Even i am scared of maths(I was a poor student :p) but I am scoring good in BCA.Basically Maths will be useful in first 2 semesters only after that there is no such need of an math expert there. Hope i have cleared your confusion.