LiIIiam0216

Let's denote each two-digit number as $10a + b,$ where $a$ is the tens digit and $b$ is the units digit. Let $n$ have $k$ two-digit numbers in its decomposition.

When we calculate the alternating sum of these two-digit numbers, we get $10a_1 + b_1 - 10a_2 - b_2 + 10a_3 + b_3 - \cdots \pm 10a_k + b_k,$ where $a_i$ is the tens digit of the $i$-th two-digit number and $b_i$ is the units digit.

This can be rewritten as $10(a_1 - a_2 + a_3 - \cdots \pm a_k) + (b_1 - b_2 + b_3 - \cdots \pm b_k).$ Let $A = a_1 - a_2 + a_3 - \cdots \pm a_k$ and $B = b_1 - b_2 + b_3 - \cdots \pm b_k.$ Then, the sum becomes $10A + B.$

Since $10A + B$ is divisible by $m,$ it implies $10A + B \equiv 0 \pmod{m},$ which gives $10A \equiv -B \pmod{m}.$

We know that $10 \equiv 1 \pmod{m},$ so $10A \equiv A \pmod{m}.$ Hence, we have $A \equiv -B \pmod{m}.$

Therefore, $n$ is divisible by $m$ if and only if $10a + b$ is divisible by $m,$ implying that the divisibility test works for $m = \boxed{11}.$

We know that a quadratic function with roots at x=2 and x=4 can be written in the factored form as:

a(x−2)(x−4)

We are also given that the function takes the value 6 when x=3. This translates to the equation: a(3−2)(3−4)=6 which simplifies to −a=6

Since we already know that a cannot be 0 (the function wouldn't be quadratic), we can divide both sides by −1 to get a=−6

Plugging this value of a back into the factored form, we get the quadratic function: −6(x−2)(x−4)

Expanding this expression, we get the answer in the desired form: −6x^2 + 36x + 48

So the answer is -6x^2 + 36x + 48.

Nevermind, I solved this.

I solved this problem!  The answer is 100*sqrt(2).

I solved this problem!  The answer is 5/2.

I solved this problem!  The answer is 12/5.

I solved this problem!  The answer is 16*sqrt(15).

I solved this problem!  The answer is 3/8.

The first term is 9 and the common ratio is 1/3, so the tenth term is 9*(1/3)^10 = 1/6561.

We can find the range of the function f(x)=2x2−8x+1 by completing the square and analyzing its extreme points.

Completing the Square:

We can rewrite the function as:

f(x)=2(x2−4x)+1

To complete the square, we take half of the coefficient of our x term, square it, and add it to both sides of the equation:

Half of the coefficient of our x term is -4 / 2 = -2.

Squaring -2 gives us 4.

Therefore:

f(x)=2(x2−4x+4)+1−8

f(x)=2(x−2)2−7

Finding the Vertex and Analyzing Extremes:

The expression inside the parentheses is squared, so it's always non-negative (0 or greater). Multiplying by 2 doesn't change this property. Therefore, the minimum value of f(x) is achieved when the squared term is 0, which happens when x=2. In this case, f(2)=2(2−2)2−7=−7.

Since the coefficient of the squared term (2) is positive, the parabola opens upwards. This means the function has a minimum point at x=2 and increases as we move away from it (either to the left or right on the interval).

Range of the Function:

The function is defined only on the interval [−1,2]. We saw that the minimum value within this interval is f(2)=−7. As x approaches either end of the interval, the function increases without bound because the squared term keeps getting larger. Therefore, the range of f(x) is the single interval:

(−7,∞)​

For f(x) to be defined, the inner term 1+x3​2​ must be defined. This inner term is undefined when its denominator, 1+x3​=0, so when x=−31​. However, even at x=−31​, the outer term 1+21​ is still defined, so f(x) is actually defined at x=−31​.

Therefore, the only restriction on the domain of f(x) comes from the denominator of the entire expression, 1+1+x3​2​. This denominator is undefined when its denominator, 1+x3​, is 0. So the function is undefined when x=−31​, as discussed above, and also when x=−1.

Finally, note that as x approaches positive or negative infinity, the entire expression approaches 11​, which is a defined value. Therefore, the only restrictions on the domain are x=−31​ and x=−1. The sum of these two numbers is −31​−1=−34​​.

We need to analyze when the expression inside the function f in g(x) becomes undefined. This happens when the denominator, (x+1), is zero. So, x=−1.

However, this restriction only applies if the numerator, (x2−2), is not also zero at x=−1. Since (−1)2−2=1=0, the function f((x2−2)/(x+1)) is defined even when x=−1.

Therefore, the only restriction on the domain of g(x) is that x=−1. This means g(x) is defined for all real numbers in the interval except for this single point.

The width of an interval that excludes a single point is 0​.

To solve this congruence, we need to find the smallest positive integer \( n \) such that \( 5^n \equiv n^5 \pmod{3} \).

First, let's observe that \( 5 \equiv 2 \pmod{3} \). Therefore, \( 5^n \equiv 2^n \pmod{3} \).

Now, let's calculate the values of \( 2^n \) modulo 3 for small values of \( n \):

- \( 2^1 \equiv 2 \pmod{3} \)

- \( 2^2 \equiv 1 \pmod{3} \)

- \( 2^3 \equiv 2 \pmod{3} \)

- \( 2^4 \equiv 1 \pmod{3} \)

From this, we can see a pattern emerge: the value of \( 2^n \) alternates between 1 and 2 modulo 3, with period 2.

Now, let's consider \( n^5 \). Since we're taking \( n^5 \) modulo 3, we can reduce \( n^5 \) to its residue modulo 3:

- \( 1^5 \equiv 1 \pmod{3} \)

- \( 2^5 \equiv 32 \equiv 2 \pmod{3} \)

- \( 3^5 \equiv 243 \equiv 0 \pmod{3} \)

The pattern for \( n^5 \) modulo 3 is not as obvious as \( 2^n \), but we can see that for \( n = 2 \), \( n^5 \) matches \( 2^n \) modulo 3.

So, the smallest positive integer \( n \) such that \( 5^n \equiv n^5 \pmod{3} \) is \( n = 2 \).