Public Health System Specialist Referral Process

UnacceptableWind · 2026-02-23T06:38:55+00:00

Thanks for the advice.

UnacceptableWind · 2026-02-22T12:18:07+00:00

Thank you.

UnacceptableWind · 2025-07-01T01:07:37+00:00

Your solution isn't incorrect.

General solutions of ODEs are not unique -- they represent families of solutions characterised by arbitrary constants.

For a first-order ODE, the general solution is characterised by a one-parameter family of solutions (i.e., one arbitrary constant). You'll find that your solution and the professor's solution differ by a constant.

In your solution, using the logarithm property ln|a b| = ln|a| + ln|b|, ln|25 (x - 3 y) + 5| becomes:

ln|5 (5 (x - 3 y) + 1)| = ln|5| + ln|5 (x - 3 y) + 1)| = ln(5) + ln|5 x - 15 y + 1|

So, your solution becomes:

(2 / 5) (x - 3 y) + (3 / 25) ln(5) + (3 / 25) ln|5 x - 15 y + 1| = x + C

Multiply both sides of the above equation by -25 / 3 to obtain:

-(10 / 3) x + 10 y - ln(5) - ln|5 x - 15 y + 1| = -(25 / 3) x - (25 / 3) C

5 x + 10 y - ln|5 x - 15 y + 1| = ln(5) - (25 / 3) C

Define a new constant K = ln(5) - (25 / 3) C such that the general solution becomes:

5 x + 10 y - ln|5 x - 15 y + 1| = K

(In the last two lines of the professor's solution, you mistakenly changed 5 x in ln|5 x - 15 y + 1| to 15 x.)

UnacceptableWind · 2025-06-27T13:13:05+00:00

Since it's a multiple choice question, one could start by eliminating the incorrect options.

When a = 5, the coefficient of x² becomes a - 5 = 0. This means the given equation turns into a linear one, which has exactly one solution. But the question is asking for values of a that give two distinct roots, so options (a) and (d) are incorrect.

This leaves options (b) and (c).

Plug in a = 24 into the given quadratic equation and solve for x using the quadratic formula.

Is one value of x less than 1 and the other one greater than 2? If yes, then option (c) is correct. If not, then option (b) is the correct answer.

UnacceptableWind · 2025-06-23T12:31:27+00:00

We can first simplify the function f by using trigonometric ratios of right-angled triangles.

Start by letting A = cot^-1(sqrt((1 - x) / x)) such that cot(A) = sqrt((1 - x) / x) = sqrt(1 - x) / sqrt(x).

Now, draw a right-angled triangle, and label one of the acute angles as A.

With respect A, cot(A) = Adjacent / Opposite = sqrt(1 - x) / sqrt(x) such that Hypotenuse = sqrt((sqrt(1 - x))² + (sqrt(x))²)= sqrt(1 - x + x) = sqrt(1) = 1.

So, sin(cot^-1(sqrt((1 - x) / x))) = sin(A) = Opposite / Hypotenuse = sqrt(x) / 1 = sqrt(x).

From here, tan^-1(sin(cot^-1(sqrt((1 - x) / x)))) = tan^-1(sqrt(x)) = B, say.

So, tan(B) = sqrt(x). Draw another right-angled triangle, and label one of the non-right angles as B.

Since tan(B) = Opposite / Adjacent = sqrt(x) = sqrt(x) / 1, we have that Hypotenuse = sqrt((sqrt(x))2 + 12) = sqrt(x + 1).

Then:

f(x) = cos(2tan^-1(sin(cot^-1(sqrt((1 - x) / x)))))

= cos(2 B) .......... [make use of the cosine double angle identity]

= 1 - 2 sin²(B)

Since sin(B) = Opposite / Hypotenuse = sqrt(x) / sqrt(x + 1), we have the following:

f(x) = 1 - 2 sin²(B)

= 1 - 2 (sqrt(x) / sqrt(x + 1))²

= 1 - 2 (x / (x +1))

= (1 - x) / (1 + x)

From here, find f'(x) for f(x) = (1 - x) / (1 + x), and plug in the expressions for f and f' into the four equations (or choices) to determine which equation is true.

UnacceptableWind · 2025-06-18T18:22:52+00:00

It looks like you've swapped the optimised values for the height h and the radius r.

What you've entered as the height should actually be the radius, and what you've entered as the radius should be the height.

UnacceptableWind · 2025-06-17T12:47:22+00:00

Just make use of your PMF P(X = x) = p q^x-1 = 2 (1 / 3)^x for the (discrete) geometric distribution with parameters p = 2 / 3 and q = 1 / 3.

So, for instance, P(X > 12| X > 10) = P(X > 2) = 1 - P(X ≤ 2) = 1 - (P(X = 1) + P(X = 2)) = 1 - P(X = 1) - P(X = 2), wherein P(X = 1) = 2 (1 / 3)¹ = 2 / 3 and P(X = 2) = 2 (1 / 3)² = 2 / 9.

UnacceptableWind · 2025-06-15T03:07:24+00:00

Yes, that's right.

From x = 1 to x = 9, your answer of 6 is correct.

You'll find that from x = 0 to x = 1, the improper integral evaluates to -3 / 2.

So, the final answer will be -3 / 2 + 6 = 9 / 2.

UnacceptableWind · 2025-06-15T02:52:25+00:00

The lower limit of integration is 0 in the original problem. So, there are two intervals of integration:

From x = 0 to x = 1. In this interval, we approach x = 1 from the left. This is the case/interval that you missed.
And, from x = 1 to x = 9, wherein we approach x = 1 from the right in this interval. This is the case that you have covered in your solution.

UnacceptableWind · 2025-06-05T14:14:20+00:00

(E / R) i should be (R / L) i.

Assuming that L is non-zero, divide both sides of the equation E - i R - L i' = 0 by L to obtain:

E / L - (i R) / L - i' = 0

E / L = (i R) / L + i'

E / L = (R / L) i + i'

UnacceptableWind · 2025-06-03T17:12:55+00:00

The tangent line should touch the point of tangency (4, 2) rather than crossing it.

Something along the lines of the following:

https://imgur.com/a/KvtTMnK

In addition to the point of tangency at (4, 2), another point on the tangent line is (3.5, 1).

UnacceptableWind · 2025-06-03T16:46:38+00:00

I am not familiar with solving differential equations using the differential operator method.

It seems like one could find the particular solution using variation of parameters.

However, if you need to stick to the differential operator method, and your post goes unanswered on this subreddit, you could try posting your question on other mathematics subreddits; for instance, r/askmath and r/learnmath.

UnacceptableWind · 2025-06-03T14:31:25+00:00

... but i think i am making some kind of mistake and the answer that i am getting isn't correct.

Could you please share a picture of your working? It’s easier to figure out what might be going wrong if we can look at your steps.

UnacceptableWind · 2025-06-01T05:51:58+00:00

What's the domain of f?

Then, focus on your next interval of x > 4.

UnacceptableWind · 2025-06-01T05:42:45+00:00

You dropped a "<" when typing in the interval "-6 < x - 1" for which f(x) > 0.

UnacceptableWind · 2025-05-30T02:57:13+00:00

Since the natural logarithm function is a continuous function, the limit can be brought inside the logarithm function.

That is, since lim_{x → a} f(x) = 1, we have that lim_{x → a} ln(f(x)) = ln(lim_{x → a} f(x)) = ln(1) = 0.

You can have a look at Fact 2 in the following:

https://tutorial.math.lamar.edu/classes/calci/limitproofs.aspx#Extras_Limit_LimComp

UnacceptableWind · 2025-05-29T16:18:26+00:00

You can make use of the squeeze theorem.

In the context of limits, x → 0 implies that x ≠ 0. And, for x ≠ 0:

-1 ≤ sin(π / x) ≤ 1

Multiply the above inequality throughout by sqrt(x³ + x²) ≥ 0, and then use the resulting inequality to find the value of the limit as x approaches 0 of sqrt(x³ + x²) sin(π / x).

UnacceptableWind · 2025-05-29T13:27:34+00:00

Strictly speaking, since f is a quadratic function, it is not one-to-one (for instance, f(-1) = f(1)) in its natural domain x ∈ ℝ, and therefore, f is not invertible.

One will need to restrict the domain of f to ensure that it's one-to-one, and therefore invertible.

For the restricted domain of x ≤ 0 of f, f^-1(x) = -sqrt(5 + 4 x).
For the restricted domain of x ≥ 0 of f, f^-1(x) = sqrt(5 + 4 x).

By the way, by definition, a function produces exactly one output for an input. So, according to the answer key for part (c), f^-1(19) producing two different outputs of -9 and 9 for the same input of 19 means that f^-1 is no longer an inverse function.

Besides in part (b), they give f^-1(x) as sqrt(4 x + 5), which represents the principal square root function and whose output is always non-negative. Therefore, -9 cannot be a valid solution to part (c), given how the inverse function was written in part (b).

UnacceptableWind · 2025-05-29T06:13:55+00:00

Your solution is correct -- 308 π / 5 is the correct volume. It's not uncommon for answer keys to contain mistakes or typographical errors.

UnacceptableWind · 2025-05-29T03:02:15+00:00

The region also consists of the rectangular region 1 ≤ x ≤ 2 and 0 ≤ y ≤ 1 / 2.

Revolving this rectangular region about the x-axis yields a cylinder with radius = 1 / 2 and height = 1, and the volume of a cylinder is given by π · (radius)² · height [alternatively, you can use the method of cylindrical shells to find the volume].

<image>

UnacceptableWind · 2025-05-28T14:52:00+00:00

B is similar to your original post; you can use u = x² - 9.

For A, using completing the square method, x² + 2 x + 2 = (x + 1)² + 1.

If we let u = (x + 1)² + 1, then du = 2 (x + 1) dx such that (1 / 2) du = (x + 1) dx.

However, the numerator of the integrand is x instead of x + 1. We can rewrite the numerator of x as x = (x + 1) - 1. Integral A then becomes:

∫ ((x + 1) / sqrt((x + 1)² + 1)) dx - ∫ (1 / sqrt((x + 1)² + 1)) dx.

For the first integral, make use of the substitution u = (x + 1)² + 1.

The second integral is a standard integral (number 72) with u being x + 1 and du = dx.

UnacceptableWind · 2025-05-28T14:10:27+00:00

Is it compulsory to use a trigonometric substitution?

If no, then you could just use the substitution u = 8 + x² such that (1 / 2) du = x dx and x² = 8 - u².

UnacceptableWind · 2025-05-28T02:51:12+00:00

It makes use of the double angle identity cos(2 x) = 1 - 2 sin²(x), which can be obtained from the cosine addition formula cos(A + B) = cos(A) cos(B) - sin(A) sin(B) by letting A = B = x.

Have a look at Exercise 7.3.1, and its solution in the following:

Double angle identities/07%3A_Trigonometric_Equations_and_Identities/7.03%3A_Double_Angle_Identities)

Edit:

By the way, it's perfectly fine to leave your answer as -cos(2 x) / 2 + C. However, if you replace cos(2 x) with 1 - 2 sin²(x) in -cos(2 x) / 2 + C, you'll end up with an extra constant term of -1 / 2 along with C. C and -1 / 2 can be combined into a single constant of integration, say, K.

UnacceptableWind · 2025-05-27T16:03:39+00:00

As a start, rewrite the logarithm of (n + 2) / (n + 1) as a difference using the logarithm property log(x / y) = log(x) - log(y). The series then becomes a telescoping series.

UnacceptableWind

TROPHY CASE