Recent content by eleventhxhour

Which value for $\theta$ is a counterexample to sin^2$\theta$+cos^2$\theta$=tan^2$\theta$ as an identity? a) pi/4 b) 5pi/4 c) pi/3 d) It is an identity So I tried subbing in each value (a, b, c) in as x and then finding the exact value from that but I'm not getting it.

If sec(2pi/3 + x) = 2, what does x equal? So far I changed it to cos by dividing 1/2. And then, I changed the 1/2 to radians which is pi/3. But, I'm not sure what to do next.

Okay, that's what I got. The textbook has it as -0.5157 so I guess it's just wrong?

Determine the average rate if change of the function y = 2cos(x - $\pi$/3) + 1 for the interval $\pi$/3 $\le$ x $\le$ $\pi$/2 I tried finding the exact values of the two (0 and 0.5) and subbing them into the AROC equation but I keep getting the wrong answer (1.4)

Given the function f(x) = 3x / (x - 2), determine the coordinates of a point on f(x) for 3 < x < 6 where the slope of the tangent line is equal to the slope of the secant line passing through A(3, 9) and B(6, 9/2). So I found that the slope of the secant line is -1.5 (therefore slope of tangent...

What are the maximum/minimum values for y = 28(1.21)^x on the interval 0 $\le$ x $\le$ 12? I think that the minimum value might be 28 because y = 28 when x = 0 but I don't know how to find the maximum value. Could someone help/explain? Thanks.

Alright, thanks! So the book also asked to graph it. I know there'd be vertical asymptote at x = -3, but how would you figure out where the horizontal asymptote is (or if there even is one)?

So I solved each of those and for the first one I got that it's negative at x<2 and x>5/2. For the second I got that it's negative at x<5/3 and x>2. Is this correct? The book had a different answer (x<5/3 and x>5/2).

So I tried that and got that it's negative for the intervals x<-3 and -3 < x < 3. But that's still different from the answer in the book? I'm not sure where they'd even get the -2 and 1/2 from.

What is the solution of |x/(x-2)| < 5 ? So, I did this the usual way of moving over the 5 to the left side and then cross multiplying and simplifying etc. However, I keep getting the wrong answer. I got x < 5/3 and x > 2, while the answer in the book says that it's x< 5/3 and x > 5/2. What...

Find the solution set for x^2/(x+3) < 9/(x+3) So I moved the term 9/(x+3) over to the left side and cross-multiplied the two fractions. Then, I simplified to get x^2-9 (because the x+3 cancel out across the fraction bar). I got x^2-9, which factors to (x+3)(x-3). Then, I created an interval...

State the range of the reciprocal function of f(x) = - (x+3)^2 - 1. I'm not sure if I did this right. I wrote that y is above/equal to -1 and below/equal to 0. Is this correct? Also, how would you graph the reciprocal function of f(x) if there is no VA and only a HA?

Solve the following inequality: 6e) $(x - 3)(x + 1) + (x - 3)(x + 2) \ge 0$ So, I created an interval table with the zeros x-3, x+1, x-3 and x+2 but I keep getting the wrong answer. Could someone help? (this is grade 12 math - so please don't be too complicated). Thanks.

So I converted each compound into moles and got 0.1025mol NaOH and 0.06molH2SO4. But I'm not sure what to do next. Would you use the equation q=mcdeltaT and find the total q for the reaction? And then use the equation deltaH=ndeltaHx? When I do this, I get the wrong answer...

Thanks! I tried that and got the right answer (: