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This is wrong

a =314

The height (in meters) of a shot cannonball follows a trajectory given by h(t) = -4.9t^2 + 14t - 0.4 at time t (in seconds).

As an improper fraction, for how long is the cannonball above a height of 6 meters?

\(\begin{array}{|rcll|} \hline -4.9t^2 + 14t - 0.4 &=& 6 \\ -4.9t^2 + 14t - 0.4 - 6 &=& 0 \quad & | \quad \\ -4.9t^2 + 14t - 6.4 &=& 0 \quad & | \quad \cdot (-1) \\ 4.9t^2 - 14t + 6.4 &=& 0 \\\\ t &=& \dfrac{14\pm \sqrt{14^2-4\cdot 4.9\cdot 6.4 } } { 2\cdot 4.9} \\\\ &=& \dfrac{14\pm \sqrt{196-125.44 } } {9.8} \\\\ &=& \dfrac{14\pm \sqrt{70.56} } {9.8} \\\\ &=& \dfrac{14\pm 8.4 } {9.8} \\\\ t_1 &=& \dfrac{14- 8.4 } {9.8} \\\\ \mathbf{t_1} & \mathbf{=} & \mathbf{ \dfrac{5.6 } {9.8} } \\\\ t_2 &=& \dfrac{14+ 8.4 } {9.8} \\\\ \mathbf{t_2} & \mathbf{=} & \mathbf{\dfrac{22.4} {9.8}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline t_2-t_1 &=& \dfrac{22.4} {9.8} - \dfrac{5.6 } {9.8} \\\\ &=& \dfrac{22.4-5.6} {9.8} \\\\ &=& \dfrac{16.8} {9.8} \\\\ &=& 1.71428571429\ \text{seconds} \\ \hline \end{array} \)

The cannonball is \(\approx 1.7\ \text{seconds} = (\dfrac{16.8} {9.8}\ \text{seconds})\)above a height of 6 meters?

If the FIRST term is 41 and the LAST term is 123, how can the "arithmetic progression" consist of " ten consecutive terms"??? In fact the "common difference" would have to be 9 1/9 !!!!!!.

delete

The mean and range of ten consecutive terms in an arithmetic progression are all 82.
What is the first term in the sequence?

\(\text{Let $a$ is the first term} \\ \text{Let $b$ is the last term}\)

\(\begin{array}{|lrcll|} \hline \text{mean:} & \dfrac{a+b}{2} &=& 82 \quad & | \quad \cdot 2 \\\\ & a+b &=& 2\cdot 82 \qquad (1) \\ \\ \text{range:} & b-a &=& 82 \qquad (2) \\ \hline \\ (1)-(2):& (a+b) - (b-a) &=& 2\cdot 82 - 82 \\ & a+b - b+a &=& 82 \\ & a+a &=& 82 \\ & 2a &=& 82 \quad & | \quad : 2 \\ & \mathbf{ a } & \mathbf{=} & \mathbf{41} \\ \hline \end{array}\)

The first term in the sequence 41

The solution of 8x+1 == 5 mod 12 is x == a mod m for some positive integers m >= 2 and a

\(\begin{array}{lrcl} &8x+1 &\equiv& 5 \pmod{12} \\ \text{or } & 8x+1 &=& 5 + 12n,\ \quad n\in Z \\\\ & 8x+1 &=& 5 + 12n \quad & | \quad -1 \\ & 8x &=& 4 + 12n \quad & | \quad :4 \\ & 2x &=& 1 + 3n \\ &&& \boxed{\mathbf{2x-3n} \mathbf{=} \mathbf{1} } \\ \end{array} \)

\(\begin{array}{|rclrcl|} \hline 2x-3n &=& 1 \quad | \quad x \text{ has the smallest coefficient} \\\\ 2x &=& 1+3n \\\\ \mathbf{x} & \mathbf{=}& \mathbf{\dfrac{1+3n}{2}} \\\\ x &=& \dfrac{1+2n+n}{2} \\\\ x &=& \dfrac{2n+(1+n)}{2} \\\\ x &=& n+\underbrace{\dfrac{ 1+n }{2}}_{=b} \qquad a \in Z \\\\ && b = \dfrac{ 1+n }{2} \\\\ && 2b = 1+n \quad | \quad n \text{ has the smallest coefficient} \\\\ && n = -1+2b \\ && \quad \quad | \quad \text{Now on the right side of the equation there is no}\\ && \quad \quad | \quad \text{break and none of the variables any more is included.} \\ && \quad \quad | \quad \text{By inserting in reverse order, now in all equations,} \\ && \quad \quad | \quad \text{in which one variable has been isolated,} \\ && \quad \quad | \quad \text{eliminates the other variables.} \\\\ x &=& \dfrac{1+3n}{2} \quad | \quad n = -1+2b \\\\ x &=& \dfrac{1+3(-1+2b)}{2} \\\\ x &=& \dfrac{1-3+6b }{2} \\\\ x &=& \dfrac{-2+6b}{2} \\\\ \mathbf{x }&\mathbf{=}& \mathbf{-1+3b} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline x & = & \underbrace{ -1 }_{=a} + \underbrace{3b}_{=m} \\\\ && \text{The positive integer $ m\ge 2$ , is $3$ or multiples of $3$}. \\ && \mathbf{m= 3,6,9,12,\ldots} \text{ and } \mathbf{a=-1} \\ \hline \end{array} \)

\(\begin{array}{|r|c|c|c|c|c|} \hline b & m=3b & m\ge 2 & x = -1+m & 8x+1 \equiv 5 \pmod {12} & x \equiv -1 \pmod{m} \\ \hline 1 & 3 & \checkmark & 2 & 17 \equiv 5 \pmod {12}\ \checkmark & 2 \equiv -1 \pmod {3}\ \checkmark \\ \hline 2 & 6 & \checkmark & 5 & 41 \equiv 5 \pmod {12}\ \checkmark & 5 \equiv -1 \pmod {3}\ \checkmark \\ \hline 3 & 9 & \checkmark & 8 & 65 \equiv 5 \pmod {12}\ \checkmark & 8 \equiv -1 \pmod {3}\ \checkmark \\ \hline \ldots & \ldots &\ldots & \\ \hline \end{array}\)

"Modulo m graph paper" consists of a grid of m^2 points,
representing all pairs of integer residues (x,y) where 0 < x < m.
To graph a congruence on modulo m graph paper, we mark every point (x,y) that satisfies the congruence.
For example, a graph of y == x^2 mod{5} would consist of the points (0,0), (1,1), (2,4), (3,4), and (4,1).
The graph of 3x == 4y-1 mod{35} has a single x-intercept (x_0,0) and a single y-intercept (0,y_0),
where 0 < x_0, y_0 < 35. What is the value of x_0+y_0?

\(\begin{array}{lrcl} &3x &\equiv& 4y-1 \pmod{35} \\ \text{or } & 3x &=& 4y-1 + 35n,\ \quad n\in Z \\\\ &&& \boxed{\mathbf{3x-4y-35n} \mathbf{=} \mathbf{-1} } \\ \end{array} \)

Solve:

\(\begin{array}{|rclrcl|} \hline 3x-4y-35n &=& -1 \quad | \quad x \text{ has the smallest coefficient} \\\\ x &=& \dfrac{-1+4y+35n}{3} \\\\ x &=& \dfrac{-1+3y+y+33n+2n}{3} \\\\ x &=& \dfrac{3y+33n+(-1+y+2n)}{3} \\\\ x &=& y+11n+\underbrace{\dfrac{ -1+y+2n }{3}}_{=a} \qquad a \in Z \\\\ && a = \dfrac{ -1+y+2n }{3} \\\\ && 3a = -1+y+2n \quad | \quad y \text{ has the smallest coefficient} \\\\ && y = 1 - 2n + 3a\\ && \qquad \qquad | \qquad \text{Since the variable n only occurs in the}\\ && \qquad \qquad | \qquad \text{integer part, set the parameter b = n. }\quad b\in Z \\\\ && \mathbf{y = 1+3a-2b} \\ && \quad \quad | \quad \text{Now on the right side of the equation there is no}\\ && \quad \quad | \quad \text{break and none of the variables any more is included.} \\ && \quad \quad | \quad \text{By inserting in reverse order, now in all equations,} \\ && \quad \quad | \quad \text{in which one variable has been isolated,} \\ && \quad \quad | \quad \text{eliminates the other variables.} \\\\ x &=& \dfrac{-1+4y+35n}{3} \quad | \quad y = 1+3a-2b,\ n=b \quad \\\\ x &=& \dfrac{-1+4(1+3a-2b)+35b}{3} \\\\ x &=& \dfrac{-1+4+12a-8b+35b}{3} \\\\ x &=& \dfrac{3+12a+27b}{3} \\\\ \mathbf{x }&\mathbf{=}& \mathbf{1+4a+9b} \\ \hline \end{array}\)

\(\begin{array}{|r|r|r|l|} \hline a & b & x = 1+4a+9b & y = 1+3a-2b & 3x \equiv 4y-1 \pmod{35} \\ \hline 1 & 2 & x_0 = 23 & 0 & 3\cdot 23 = 4\cdot0-1 \pmod{35} \\ & & 0\lt 23 \lt 35 \ \checkmark & & 69 = -1 \pmod{35} \ \checkmark \\ \hline 2 & -1 & 0 & y_0 = 9 & 3\cdot0 = 4\cdot 9 -1 \pmod{35} \\ & & & 0\lt 9 \lt 35 \ \checkmark & 0 = 35 \pmod{35} \\ & & & & 0 = 35-35 \pmod{35} \\ & & & & 0 = 0 \pmod{35} \ \checkmark \\ \hline \end{array}\)

\(x_0+y_0 = 23+9=32\)

Ask on the AoPS Message Board - the instructor will help....

SOMEBODY PLEASE HELP

2.51^x must be positive so 3-2.52^x <3

So the range (-infinity, 3)

check

\(f(x) = 3- 2\cdot 51^x\)

Since each student membership costs $25, 25 per month and x students were members that month, we have that 25x, x represents the money collected from student memberships.

Similarly, since each adult membership costs $40, 40 per month, and y adults were members that month, we have that 40y, y represents the money collected from adult memberships.

figure the rest out

A variation on this question has come up before on this forum. The most important example is discussed here in detail: https://web2.0calc.com/questions/probability_882

Lightning, you are NOT stupid so how about displaying your questions properly!

A lily pad doubles in size each day. If it takes 28 days for the lily pad to cover the entire pond, how many days will it take to cover one-fourth of the pond?

T0=1

T1=2

T2=4

Tn=2^n

T28=2^28 

find n such that  Tn=(2^28)/4 = 2^26

26 days. Assuming that there is just 1 pad to start with.

I'd really like a mathematician to talk through this question/answer.     

What function?

The range is the set of all possible y values.

By simple inspection, it turns out that p=2 and p=3, for which there is no integer solution in x for the congruence 3(6x+1) == 4 mod(p).

Their sum = 2 + 3 = 5

There is a formula you can use to calculate the probability of all faces showing on any number of dice. From that, you can easily determine the minimum number of dice you need to roll. Here is the formula in question: Just change "n" as required and it will give you the probability.

n = 10;  1 - ((6C1) *(5/6)^n - (6C2)*(4/6)^n + (6C3)*(3/6)^n - (6C4)*(2/6)^n + (6C5)*(1/6)^n).

Here are half a dozen calculations to give you an idea: 

n=10        p=0.271 812 128 5

n=11        p=0.356 206 418 6

n=12        p=0.437 815 680 6

n=13       p=0.513 858 194 0

n=14       p=0.582 845 348 9

n=15       p=0.644 212 738 6

As you can easily see that you will need to roll a minimum of 13 dice in order to get at least 50% probability that all 6 faces will show.

The total volume is 27 so each side has length 3. Boring a single hole through the cube removes a region of size , so removes a volume of 3, however these volumes overlap in the center  region of the cube.

The first hole drilled removes a volume of 3. The second and third hole drilled only remove a volume of 2, so the total resulting volume is
27-3-2-2=20

so your answer is 20

See https://web2.0calc.com/questions/polynomial_6#r1