GYanggg

Hey Guest

We first need to factor the public modulus \(M = pq = 527 \Rightarrow 527 = 17 \cdot 31. \)

Therefore \(p =17 \ \&\ q=31 \)

The definition of a secret decryption exponent is the multiplicative inverse of the public encoding exponent, in modulo \( (p-1)(q-1)\).

Therefore \(d= 37^{-1} \pmod{480}. \Rightarrow d=\boxed{13}\)

I hope this helped,

Gavin

If the roots of the quadratic equation \(\frac12x^2+99x+c=0 \) are \(x=-99+\sqrt{8001}\) and \(x=-99-\sqrt{8001},\) then what is the value of \(c\)?

\(c=\boxed{900}\)

I hope this helped,

Gavin

We can model the rate of a change as a right triangle with base \(x = 400\) feet and height y increasing at a rate of 10 feet per second.

After 30 seconds, \(y=10\cdot30=300\)

If the distance between Eric and Harrison is z, the Pythagorean Theorem gives us: \(z^2 = x^2 + y^2 \).

The distance between Eric and Harrison is: \(x^2=300^2+400^2\Rightarrow x=500\)

Differentiating both sides also yields \(2z · z' = 2x · x' + 2y · y'\). Plugging in our values, we get \(2 · 500z' = 2 · 400 · 0 + 2 · 300 · 10\), which gives us \( z' = 6\) feet per second.

I hope this helped,

Gavin

When the rectangle perfectly covers the curved surface of the cylinder without overlapping itself at all, the length of the rectangle forms a circle that the the base of the cylinder. 

When we know the area of the base of the cylinder, we can multiply it by the height to find the volume. 

Since the length of the rectangle is 24, the radius of the circle is \(\frac{12}{\pi}\). 

The area is: 

\((\frac{12}{\pi})^2\cdot\pi=\frac{144}{\pi}\)

The volume is \(144\cdot8\div\pi\)

I hope this helped,

Gavin

\(\text{midline}=\frac{\text{max value}+\text{min value}}{2}\)

Since the amplitude of a sinusoid is equal to the vertical distance from its midline to one of its extrema, we can also calculate the midline using one of the following formulas.

\(\text{midline}=\text{max value}-\text{amplitude}\)

\(\text{midline}=\text{min value}+\text{amplitude}\)

I'm going to leave the calculating to you. 

I hope this helped,

Gavin

\(i^{-100}+i^{-99}+i^{-98}+\cdots+i^{-1}+i^0+i^1+\cdots+i^{99}+i^{100}\)

\(i=\sqrt{-1}, \text{ we start simple and try to find a pattern.}\)

\(i^0=1\\ i^1=i\\ i^2=-1\\ i^3=-i\\ i^4=1\\ i^5=i\\ i^6=-1\\ i^7=-i\\ \)

Every 4 terms, the sequence rotates, \(1, i , -1, -i\)

Their sum is: \(1+i-1-i=0\)

Since every four terms cancel out, we just need to know what term the sequence starts and ends. 

\(i^{-100}=1\\ i^{100}=1\\\)

\(i^{-100}+i^{-99}+i^{-98}+\cdots+i^{-1}+i^0+i^1+\cdots+i^{99}+i^{100}\\ =1+i-1-i+1+-\cdots+1+i-1-i+1\\ =\boxed1\)

I hope this helped,

Gavin

Hey FiestyGeco!

I can only help you with 3, since the pictures won't show. 

For 3, we use \(c^2=a^2+b^2-2ab\cos\gamma\), the law of cosines.

This formula generalizes the Pythagorean Theorem, since \(\cos90º=0\Rightarrow a^2+b^2=c^2\)

We can write: \(CB^2=CA^2+AB^2-2(AC)(AB)-\cos{A}\)

\((4\sqrt3)^2=(4\sqrt2)^2+AB^2-2(4\sqrt2)(AB)-\cos60\\ 48=32+AB^2-8\sqrt2AB-\frac12\\ \)

After solving for AB, you use the same formula to solve for the angle.         

This doesn't look like the best method, does anyone know a less rigid way of doing this?

I hope this helped,

Gavin

Since lines AB aned DE are parallel, angle ABC is equal to angle BCE. 

Therefore, ACB = 180 - 50 - 80 = 50

Since ACB = 50, BAC = 180 - 80 - 50 = 50 

The three angles of the triangle are 50, 50, 80. 

This means that the triangle is isosceles.

\(\text{Compute}:\displaystyle\lim_{x\to0}(\frac{\sin{x}}{x})^\frac{1}{1-\cos{x}}\)

We can approximate \(\sin{x} \text{ and } \cos{x}\) by their Taylor series.

After apply substitutions:

\(\displaystyle\lim_{x\to0}(\frac{\sin{x}}{x})^\frac{1}{1-\cos{x}}=\lim_{x\to0}(1-\frac{x^2}{6})^\frac{2}{x^2}=\lim_{x\to0}(1+\frac{x^2}{6})^{-\frac{2}{x^2}}\)

\(=\displaystyle\lim_{x\to\infty}(1+\frac{1}{6x^2})^{-2x^2}=\lim_{x\to\infty}(1+\frac{1}{x})^{-x/3}=\boxed{e^{-1/3}}\)

This is because \(e=\displaystyle\lim_{x\to\infty}(1+\frac1x)^x\)

I hope this helped,

Gavin

\(\text{Find } a: \int_1^a(3x^2-6x+3)dx=27\)

\(\text{The solution you seek is } a=\boxed4\\ \)