Author Topic: Mathematics Question Thread (Read 1303504 times) Tweet Share

RuiAce · « **Reply #840 on:** October 23, 2016, 04:05:14 pm »

Quote from: katnisschung on October 23, 2016, 03:40:02 pm

thanks ruiAce

I've attached it here

$\text{The information in Q6 has been annotated over this diagram.}\\ \text{It remains to show that }\angle OAN_1=30^\circ\\ \text{to show that }\angle OAC = 60^\circ$
$\text{As you have stated, }\angle OAN_1=30^\circ\text{ by co-interior angles}$
And then part iii) can be done by using the angle sum of a triangle, and the sine rule.

asd987 · « **Reply #841 on:** October 24, 2016, 06:40:12 pm »

How would I integrate x^-3 / 3

jakesilove · « **Reply #842 on:** October 24, 2016, 07:11:10 pm »

Quote from: asd987 on October 24, 2016, 06:40:12 pm

How would I integrate x^-3 / 3

Hey! To integrate something like that, you just use the normal anti-derivative formula

$\int{x^n}dx=\frac{x^{n+1}}{{n+1}}+C$

Apply this directly to the equation you've given us! You can bring the 1/3 outside the integral

asd987 · « **Reply #843 on:** November 01, 2016, 09:57:36 pm »

hi i've got a question.
the area bounded by the curve y=x^2 and y=x+2 is rotated about the x axis. find the exact volume of the solid formed

RuiAce · « **Reply #844 on:** November 01, 2016, 10:05:12 pm »

Quote from: asd987 on November 01, 2016, 09:57:36 pm

hi i've got a question.
the area bounded by the curve y=x^2 and y=x+2 is rotated about the x axis. find the exact volume of the solid formed

$\text{From sketching the curves, clearly the line }y=x+2\\ \text{is above the curve }y=x^2$
$\text{We may identify the points of intersection by solving }x^2=x+2\\ \text{The answers will be }x=-1, x=2$
$\text{So just like how the area between two curves is }\int_a^b (y_1-y_2)dx\\ \text{Using the volume between two curves }V=\pi\int_a^b(y_1^2-y_2^2)dx$
$\begin{align*}V&=\int_{-1}^2((x^2)^2 - (x+2)^2)dx\\ &= \int_{-1}^2(x^4-x^2-4x-4)dx\\ &=\left[\frac{x^5}{5}-\frac{x^3}{3}-2x^2-4x\right]_{-1}^2\end{align*}$

Aussie1Italia2 · « **Reply #845 on:** November 03, 2016, 12:08:05 pm »

Hey, I have a series of question I need help with; please and thank you.

1. Find the discriminant of the quadratic equation:
A) -x² - 3x = 0
B) x² + 4 = 0

2. Find any values of b for which 2x² + x + b + 1 = 0

Again, thanks!!!

RuiAce · « **Reply #846 on:** November 03, 2016, 12:12:42 pm »

Quote from: Aussie1Italia2 on November 03, 2016, 12:08:05 pm

Hey, I have a series of question I need help with; please and thank you.

1. Find the discriminant of the quadratic equation:
A) -x² - 3x = 0
B) x² + 4 = 0

2. Find any values of b for which 2x² + x + b + 1 = 0

Again, thanks!!!

$\text{That's just using the formula }\Delta = b^2-4ac\\ \text{For A) you have }a=-1, \, b=-3, \, c=0\\ \text{So }\Delta =9$

Your second question is inconclusive. Because the way you put it, it depends on your value for x. Suspecting something in the question wasn't typed because I can sort of tell what the question should be.

Aussie1Italia2 · « **Reply #847 on:** November 03, 2016, 12:42:06 pm »

Oh yep thanks;

I forgot the last bit of the question.

Find any values of b for which 2x² + x + b + 1 = 0 has real roots.

jamonwindeyer · « **Reply #848 on:** November 03, 2016, 12:44:04 pm »

Quote from: Aussie1Italia2 on November 03, 2016, 12:42:06 pm

Oh yep thanks;

I forgot the last bit of the question.

Find any values of b for which 2x² + x + b + 1 = 0 has real roots.

Cool! For real roots we require our discriminant to be greater than or equal to zero!

$\Delta\ge0\\b^2-4ac\ge0\\1^2-4(2)(b+1)\ge0\\-8b-7\ge0\\b\le\frac{-7}{8}$

RuiAce · « **Reply #849 on:** November 03, 2016, 12:45:45 pm »

$\text{Recall that the condition for real roots is that }\Delta \ge 0\\ \text{If }\Delta < 0\text{, then we have no real roots.}\\ \text{Whereas otherwise, 1 or 2 roots are real.}$
$\Delta = (1)^2 - 4(2)(b+1) = 1-8(b+1) = -8b-7$
$\text{So }\Delta \ge 0\implies -8b-7 \ge 0 \implies b \le- \frac78$

Quote from: jamonwindeyer on November 03, 2016, 12:44:04 pm

Cool! For real roots we require our discriminant to be greater than or equal to zero!

$\Delta\ge0\\b^2-4ac\ge0\\1^2-4(2)(b+1)\ge0\\-8b-7\ge0\\b\le\frac{-7}{8}$

Well damn I didn't type fast enough

Hplovers · « **Reply #850 on:** November 03, 2016, 03:36:23 pm »

Would be great if you could help me out with 4 (ii) please

Thankyou!

RuiAce · « **Reply #851 on:** November 03, 2016, 03:44:48 pm »

Quote from: Hplovers on November 03, 2016, 03:36:23 pm

Would be great if you could help me out with 4 (ii) please
Thankyou!

$\text{Given }\triangle EBF ||| \triangle AED\text{ from the proportional sides we have}\\\frac{EF}{FB}=\frac{AD}{DE}\\ \text{But }CFED\text{ is a rhombus, so }CF=FE=ED=DC=x$
$\text{So subbing it in we have}\\ \begin{align*}\frac{x}{a-x}&=\frac{b-x}{x}\\ \implies x^2&=(b-x)(a-x)\\ x^2&=ab-(a+b)x+x^2\\ \implies (a+b)x&=ab\\ \therefore x&=\frac{ab}{a+b}\end{align*}$

Hplovers · « **Reply #852 on:** November 03, 2016, 03:55:45 pm »

Quote from: RuiAce on November 03, 2016, 03:44:48 pm

$\text{Given }\triangle EBF ||| \triangle AED\text{ from the proportional sides we have}\\\frac{EF}{FB}=\frac{AD}{DE}\\ \text{But }CFED\text{ is a rhombus, so }CF=FE=ED=DC=x$
$\text{So subbing it in we have}\\ \begin{align*}\frac{x}{a-x}&=\frac{b-x}{x}\\ \implies x^2&=(b-x)(a-x)\\ x^2&=ab-(a+b)x+x^2\\ \implies (a+b)x&=ab\\ \therefore x&=\frac{ab}{a+b}\end{align*}$

Thankyou so so much!

teapancakes08 · « **Reply #853 on:** November 04, 2016, 09:43:08 am »

Probably should be able to answer this, but the second term is throwing me off:

4. Simplify:
(h) 1 - sin^2 (180-x)

If it was just 1-sin^2x it would be cos^2x, but...

If someone could help clarify that would be great, thanks ^^

RuiAce · « **Reply #854 on:** November 04, 2016, 09:55:49 am »

Quote from: teapancakes08 on November 04, 2016, 09:43:08 am

Probably should be able to answer this, but the second term is throwing me off:

4. Simplify:
(h) 1 - sin^2 (180-x)

If it was just 1-sin^2x it would be cos^2x, but...

If someone could help clarify that would be great, thanks ^^

$\text{Note that }\sin(180^\circ-x)=\sin x\quad \text{(second quadrant angle)}$

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