Recent content by suluclac

With the possibility that some school students come in this forum and expect us to spoon feed the answers, here it is. Besides, one of the rules of this forum is that we need to show some effort! So here is my effort!

If a line intersects both points (2, 2) and (6, 7), determine its slope.

Correct.

$$\frac{3(-3)-(5)2^2}{8-\sqrt{36}+12}=\frac{3(-3)-(5)4}{8-6+12}=\frac{-9-20}{8-6+12}=-\frac{29}{14}$$

It could be. If so, what's the next step? What does the sine squared equal to?

I'll take that as a no.

Does the problem from the book say 4((a² - b²)cd + ab(c² - b²))² + ((a² - b²)(c² - b²) - 4abcd)²?

This problem is similar to this. I tried using the theorem of Pappus. Is $$\pi/6$$ incorrect? How did you get your answer?

$$\frac{1}{4}-\frac{1}{4}\left(\frac{\text{?}}{\text{?}}\right)$$ What Mark is saying here "deal with it in the same way" is applying the simple trig formula for $$\cos^2{2x}$$. WLOG, $$\cos^2{u}=\frac{1+\cos{2u}}{2}$$.

Hello, I don't see a square root in the expression. Are you asking us how to factor the expression?

The region between y = x & y = -x² + 2x revolves around y = x. Determine the volume.

If this was on the exam, some students could multiply 2x + 7 itself 8 times and get the desired result. Hopefully, the student has studied the binomial theorem before the exam!

Converges absolutely

$$-\int\cos^3{2x}\,dx=-\frac{1}{4}\int\cos{6x}\,dx-\frac{3}{4}\int\cos{2x}\,dx=-\frac{1}{4}(\frac{1}{6})\sin{6x}-\frac{3}{4}(\frac{1}{2})\sin{2x}+C$$ $$\therefore-\int\cos^3{2x}\,dx=-\frac{1}{24}\sin{6x}-\frac{3}{8}\sin{2x}+C$$ This, of course, requires the reader the ability to prove...