Any calculator and math genius's out there? i have some tricky math problems?

Question

1. On a calculator with a square-root, square,reciprocal, trignometric, and inverse-trigonometric keys, the value of x is in the display. Evaluate the expression... x/(sqrt(1-x^2)) by pressing only two or three keys (once each).


2. The "adderage" of two fractions is defined as the sum of their numerators over the sum of their denominators. When, if at all, will the adderage of two fractions equal the average of the same two fractions?

3. What's the largest whole number that must be a factor of n^5-5n^3+4n for all whole numbers n?

4. solve the system x^2+y^2=208 and xy=96
or this system x+y+xy=19 and xy(x+y)=84

5. D and M were one mile apart when they began walking toward each other. D walks at a constant rate of 3 MPH and M walks at a constant rate of 4MPH. When they started, D's dog, who ran at a constant rate of 7MPH, ran to m; then turned back and ran back and forth until D and M met and lost no time turning around, how far did it run?

Thanks everyone!

Answer 1

1. In order for sqrt(1-x²) to be a real number, x must be between -1 and 1. In this case, you can compute the expression with these keystrokes:

[inv] [sin] [tan]

Since x is a real number between -1 and 1, we can substitute sinθ for x. When we apply the identity sin²θ + cos²θ = 1, the expression x/(sqrt(1-x²)) becomes sinθ/cosθ, that is, tanθ.

So [inv] [sin] computes the angle whose sine is 'x', and [tan] computes tangent of that angle, which is equal to the desired result.

2. The 'adderage' will always always equal the average if the 2 denominators are the same, i.e. 1/7 and 4/7, and also when both fractions express the same value, i.e. 1/2 and 2/4 or 3/6 and 4/8. In the more general case, if the fractions are a/b and c/d, then it's true whenever ad² + bc² = abd + bcd. This is an example of a diophantine problem: the number of equations is less than the number of unknowns and the solutions must be integers.

3. The answer is 120.

For whole-number values of n less than or equal to 2, the expression evaluates to zero. I'll assume we're only considering values of n > 2.

Factoring n^5-5n^3+4n, we get (n+2)(n+1)(n)(n-1)(n-2). That's five numbers in descending sequence. Such a sequence will always contain at least one multiple of '5', one multiple of '4', one multiple of '3' and one multiple of '2', so the product must always be divisible by 5 x 4 x 3 x 2, that is, 120.

4a. since xy=96, you can add '2xy=192' to both sides, giving the factorable: x² + 2xy + y² = 208 + 192, or (x+y)² = 400, thus
|x+y| = 20, that is, x+y=20 or x+y=-20. Taking x+y=20, y=20-x so x(20-x)=96; 20x - x² = 96; x²-20x+96=0; roots at 8 and 12; then because of the absolute value, you also have roots at -8 and -12. The full solution set is:
(x,y) ∈{(8,12), (12,8), (-8,-12), (-12,-8)}

4b. since x+y+xy=19, we know x+y=19-xy. We sub '19-xy' for 'x+y' into xy(x+y)=84 and get: xy(19-xy)=84. Expand and factor: (xy)² - 19xy + 84 = 0 --> (xy-12)(xy-7)=0. This has roots at (xy)=12 and (xy)=7. Substituting back into the original equation and solving via the quadratic equation, we obtain the complete solution set:
(x,y) ∈{(3,4), (4,3), (6-√29,6+√29), (6+√29,6-√29)}

5. 1 Mile. D and M have a combined speed of 7 MPH, the same speed as the dog. Thus, when D and M meet, they will have together covered 1 mile and the dog, running at the same speed of 7 MPH, will have gone the same distance.

Answer 2

Let me take a whack at nos. 3 and 4.
3. A trick question?
There is no such largest whole number
because n always divides n^5-5n³+4n !

4. x² + y² = 208
xy = 96
(x+y)² = x²+2xy+y² = 208+192 = 400
So x+y = 20 or -20.
Suppose x+y=20.
Then y = 96/x.
x +96/x = 20.
x²-20x+96=0.
(x-12)(x-8)=0
x=12, y = 8
or x = 8, y = 12.
Similarly, if x+y= -20
we find
x = -12,y= -8
or x= -8, y = -12.

Answer 3

all of us be attentive to that 2 numbers sum to 10 (upload to 10) and in addition they multiply to 5. permit's say the two numbers are: x and y consequently: x + y = 10 xy = 5 resolve utilising our gadget of equations suggestions x = 10 - y substitute into 2d equation (10-y)y = 5 10y - y^2 = 5 0 = y^2 - 10y + 5 utilising the quadratic formula, you will get: y = 5 +/- 2sqrt(5) properly then x may be the two: x = 10 - y x = 10 - (5 + 2sqrt(5)) or x = 10 - (5 - 2sqrt(5)) so our solutions are this: x = 5 - 2sqrt(5), y = 5 + 2sqrt(5) OR x = 5 + 2sqrt(5), y = 5 - 2sqrt(5) voila