Author Topic: Mathematics Question Thread (Read 1304230 times) Tweet Share

RuiAce · « **Reply #1875 on:** May 21, 2017, 11:17:21 pm »

Quote from: Fahim486 on May 21, 2017, 10:59:54 pm

Hey, I'm having trouble doing these questions because I suck at trigonometry.
Thanks!

$\begin{align*}1-\sin^2\left(\frac{\pi}{2}-\theta\right)&=1-\cos^2\theta\\&=\sin^2\theta\end{align*}$
$\tan^2x - 1 = 0 \implies \tan x = \pm 1$
$\text{Similar to a recently asked question we need to consider all four quadrants}\\ \therefore x= \frac{\pi}4, \pi - \frac{\pi}{4}, \pi + \frac{\pi}{4}, 2\pi -\frac{\pi}4$

RuiAce · « **Reply #1876 on:** May 21, 2017, 11:24:32 pm »

Quote from: JuliaPascale123 on May 21, 2017, 09:37:41 pm

(Image removed from quote.)

$\text{From }y=\sin x \text{ we first sketch }y=\sin \frac{\pi x}{4}\\ \text{by changing the period to 8}\\ \text{Then, shift the graph to the right by 1 unit to obtain }y=\sin \frac{\pi(x-1)}{4}$
$\text{Adjust the amplitude to be 3 to obtain }y=3\sin \frac{\pi(x-1)}{4}\\ \text{And shift the graph up by 4 to obtain }y=4+\sin \frac{\pi(x-1)}{4}\\ \text{making 4 the 'average' value}$
You can sketch the graph on software such as Desmos
_____________-
$\text{You are sketching }H=60-20\cos \frac{\pi t}{6}$

Rasika · « **Reply #1877 on:** May 22, 2017, 01:15:41 am »

What's the best way to understand a hard concept?

jamonwindeyer · « **Reply #1878 on:** May 22, 2017, 01:19:13 am »

Quote from: Rasika on May 22, 2017, 01:15:41 am

What's the best way to understand a hard concept?

Lots of practice! Maths is a skill best learned by putting your brain to work and just trying the question type you are struggling with over and over, having someone step you through initially and then taking the reigns yourself. Of course there are other things you can do, but it all ultimately comes back to practice makes perfect

Athos36 · « **Reply #1879 on:** May 22, 2017, 10:37:14 am »

Hi there! (once again) Could you help me with a quadratic question please?

A parabola has form y=ax^2+bx+4.
a) its turning point is on the x-axis. Find "a" in terms of "b" (answer= b^2/16)
b) If turning point is at (-4,0), find values "a" and "b". )answer= a= 1/4; b= 2)

Thanks in advance!

RuiAce · « **Reply #1880 on:** May 22, 2017, 10:41:27 am »

Quote from: Athos36 on May 22, 2017, 10:37:14 am

Hi there! (once again) Could you help me with a quadratic question please?

A parabola has form y=ax^2+bx+4.
a) its turning point is on the x-axis. Find "a" in terms of "b" (answer= b^2/16)
b) If turning point is at (-4,0), find values "a" and "b". )answer= a= 1/4; b= 2)

Thanks in advance!

$\text{The turning point is at }x=-\frac{b}{2a}\text{ by formula}\\ \text{Alternatively, prove it very quickly with calculus}\\ y=ax^2+bx+4\implies \frac{dy}{dx}=2ax+b.\text{ Setting it equal to 0 gives }x=-\frac{b}{2a}$
$\text{In part a), we are told that when }x=-\frac{b}{2a}, y=0\\ \therefore 0 = a\left(\frac{b}{4a^2}\right)+b\left(-\frac{b}{2a}\right)+4\\ \text{Which rearranges to yield the desired result}$
b) is now taking $x=-\frac{b}{2a}=4$

Jake accidentally edited my post

As usual, Rui beats me to the punch. 'Different' methods, though, so hopefully it's helpful!

jakesilove · « **Reply #1881 on:** May 22, 2017, 10:42:29 am »

Quote from: Athos36 on May 22, 2017, 10:37:14 am

Hi there! (once again) Could you help me with a quadratic question please?

A parabola has form y=ax^2+bx+4.
a) its turning point is on the x-axis. Find "a" in terms of "b" (answer= b^2/16)
b) If turning point is at (-4,0), find values "a" and "b". )answer= a= 1/4; b= 2)

Thanks in advance!

Hey! We know that the turning point is on the x-axis, which occurs when y=0. Therefore,
$0=ax^2+bx+4$
Now, we can use the quadratic formula to solve this equation!
$x=\frac{-b\pm \sqrt{b^2-16a}}{2a}$

However, we know that there can only be one solution to this! Think about it; if a parabola has a turning point on the x-axis, it can't have more than one x-intercept. Therefore, the square root must be equal to zero (do you understand why?)
$b^2-16a=0$
$a=\frac{b^2}{16}$

As required
Now, we know that the turning point must occur at
$x=\frac{-b}{2a}$

To answer this part of the question, sub in our value of 'a' that we found in the first part of the question, and set x=-4. Then, solve for b, and for a! Let me know if you need more help with this part

Welcome to the forum!

1937jk · « **Reply #1882 on:** May 22, 2017, 08:52:35 pm »

Hey, just really stuck on this question,
The sum of 50 terms of an arithmetic series is 249 and the sum of 49 terms of the series is 233. Find the 50th term of the series.

RuiAce · « **Reply #1883 on:** May 22, 2017, 08:59:23 pm »

Quote from: 1937jk on May 22, 2017, 08:52:35 pm

Hey, just really stuck on this question,
The sum of 50 terms of an arithmetic series is 249 and the sum of 49 terms of the series is 233. Find the 50th term of the series.

$\text{All the ingredients are given to you}\\ \begin{align*}S_{49} + T_{50}& = S_{50}\\ 233 + T_{50} &= 249\\ T_{50}&=16\end{align*}$
$\text{Use the definition and hence your intuition}\\ S_n\text{ is just }S_{n-1}\text{ plus the next term in the sequence}$

RuiAce · « **Reply #1884 on:** May 24, 2017, 10:54:59 am »

Quote from: jakesilove on May 22, 2017, 10:42:29 am

Hey! We know that the turning point is on the x-axis, which occurs when y=0. Therefore,
$0=ax^2+bx+4$
Now, we can use the quadratic formula to solve this equation!
$x=\frac{-b\pm \sqrt{b^2-16a}}{2a}$

However, we know that there can only be one solution to this! Think about it; if a parabola has a turning point on the x-axis, it can't have more than one x-intercept. Therefore, the square root must be equal to zero (do you understand why?)
$b^2-16a=0$
$a=\frac{b^2}{16}$

As required
Now, we know that the turning point must occur at
$x=\frac{-b}{2a}$

To answer this part of the question, sub in our value of 'a' that we found in the first part of the question, and set x=-4. Then, solve for b, and for a! Let me know if you need more help with this part

Welcome to the forum!

$\text{There was a query regarding how to solve for }a\text{ and }b$
$\text{Upon setting }x=4\text{, in the second equation we have}\\ \boxed{4=-\frac{b}{2a}}\\ \text{which rearranges to }\boxed{8a=-b}$
$\text{In the equation from the previous part we also had }\boxed{a=\frac{b^2}{16}}\\$
$\text{This is now a pair of simultaneous equations}\\ \text{By substitution, we have }a=\frac{64a^2}{16} \implies a = 4a^2\\ \text{This is now a quadratic equation }4a^2-a=0\text{ which has solutions }a=0,\frac{1}{4}$
$\text{From here, figure out why }a=0\text{ should be rejected}\\ \text{and then it's easy to find }b$

Athos36 · « **Reply #1885 on:** May 24, 2017, 11:11:32 am »

I'm just finding it difficult to teach myself. You are way ahead of me since my math knowledge is rather small. I'll just have to muddle through the problem.
Thanks for your time.

RuiAce · « **Reply #1886 on:** May 24, 2017, 11:34:26 am »

What part needs further addressing?

jamonwindeyer · « **Reply #1887 on:** May 24, 2017, 02:49:26 pm »

Quote from: Athos36 on May 24, 2017, 11:11:32 am

I'm just finding it difficult to teach myself. You are way ahead of me since my math knowledge is rather small. I'll just have to muddle through the problem.
Thanks for your time.

Definitely let us know which bit doesn't quite click. I know it's frustrating not quite getting it right away, but we're happy to go into more detail where you need it

JuliaPascale123 · « **Reply #1888 on:** May 24, 2017, 04:16:01 pm »

Quote from: RuiAce on May 21, 2017, 11:24:32 pm

$\text{From }y=\sin x \text{ we first sketch }y=\sin \frac{\pi x}{4}\\ \text{by changing the period to 8}\\ \text{Then, shift the graph to the right by 1 unit to obtain }y=\sin \frac{\pi(x-1)}{4}$
$\text{Adjust the amplitude to be 3 to obtain }y=3\sin \frac{\pi(x-1)}{4}\\ \text{And shift the graph up by 4 to obtain }y=4+\sin \frac{\pi(x-1)}{4}\\ \text{making 4 the 'average' value}$
You can sketch the graph on software such as Desmos
_____________-
$\text{You are sketching }H=60-20\cos \frac{\pi t}{6}$

Thanks so much.

What would the answer for b) be?

http://i68.tinypic.com/10ynmkn.png

1937jk · « **Reply #1889 on:** May 24, 2017, 04:16:46 pm »

Quote from: RuiAce on May 22, 2017, 08:59:23 pm

$\text{All the ingredients are given to you}\\ \begin{align*}S_{49} + T_{50}& = S_{50}\\ 233 + T_{50} &= 249\\ T_{50}&=16\end{align*}$
$\text{Use the definition and hence your intuition}\\ S_n\text{ is just }S_{n-1}\text{ plus the next term in the sequence}$

Okay cool! Thank you I see now!

Another question haha,
Which term of the series 8, -4 , + 2.... is 1/128?

I'm just particularly struggling with dealing with a negative r value and using log to solve for n
thanks!

Mod Edit [Aaron]: Merged double post. Please edit your previous post if you want to add something.

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