All Answers 352964 Answers

Let \(f(x)=3x+2\) and \(g(x)=ax+b\), for some constants \(a\) and \(b\).
If \(ab=20\) and \(f(g(x))=g(f(x))\) for \(x=0,1,2,\ldots ,9\), find the sum of all possible values of \(a\).

Let \(f(x)=3x+2\) and \(g(x)=ax+b\), for some constants \(a\) and b.
If \(ab=20\) and \(f(g(x))=g(f(x))\) for \(x=0,1,2,\ldots ,9\), find the sum of all possible values of \(a\).

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{3x+2} \quad &| \quad x=g(x) \\\\ f(g(x)) &=& 3g(x)+2 \quad &| \quad g(x)=ax+b \\ f(g(x)) &=& 3(ax+b)+2 \\ f(g(x)) &=& 3ax+3b+2 \quad &| \quad b=\dfrac{20}{a} \\ f(g(x)) &=& 3ax+3*\dfrac{20}{a}+2 \\ \mathbf{f(g(x))} &=& \mathbf{3ax+ \dfrac{60}{a}+2} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{g(x)} &=& \mathbf{ax+b} \quad &| \quad x=f(x) \\\\ g(f(x)) &=& af(x)+b \quad &| \quad f(x)=3x+2 \\ g(f(x)) &=& a(3x+2)+b \\ g(f(x)) &=& 3ax+2a+b \quad &| \quad b=\dfrac{20}{a} \\ \mathbf{g(f(x))} &=& \mathbf{3ax+2a+\dfrac{20}{a}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{ f(g(x)) } &=& \mathbf{g(f(x))} \\\\ 3ax+ \dfrac{60}{a}+2 &=& 3ax+2a+\dfrac{20}{a} \\\\ \dfrac{60}{a}+2 &=& 2a+\dfrac{20}{a} \\\\ \dfrac{60}{a}-\dfrac{20}{a}+2 &=& 2a \\\\ \dfrac{40}{a} +2 &=& 2a \quad &| \quad * a \\\\ 40 +2a &=& 2a^2 \quad &| \quad : 2 \\ 20 + a &=& a^2 \\ \mathbf{ a^2-a-20} &=& \mathbf{0} \\\\ a _{1,2} &=& \dfrac{1 \pm \sqrt{1-4(-20)} }{2} \\ a _{1,2} &=& \dfrac{1 \pm \sqrt{81} }{2} \\\\ a _{1,2} &=& \dfrac{1 \pm 9 }{2} \\\\ a_1 = \dfrac{1}{2} + \dfrac{9}{2} && a_2 = \dfrac{1}{2} - \dfrac{9}{2} \\ a_1+a_2 &=& 2\times \dfrac{1}{2} \\ \mathbf{a_1+a_2} &=& \mathbf{1} \\ \hline \end{array}\)

Thanks Tertre, that was so obvious .. after you explained it     

The solutions are z = 1 + i, 1.5890 - 1.2877i, 1.5890 + 1.877i.

1. By power of a point, the radius is sqrt(33).

2. by power of a point, CM = 5*sqrt(3).

You can check geno's answer here:

x/13=69

x=69*13

x=897

https://web2.0calc.com/questions/reposted-again-bc-a-certain-user-kept-complaining#r7

OMG THANKYOU

I believe I posted an answer to your first question already. You can find the solution here:

https://web2.0calc.com/questions/reposted-again-bc-a-certain-user-kept-complaining#r7

let me know if you see anything wrong with my solution, and I'll do my best to correct it.

Also, don't repost your question after it hasn't been answered for an hour or two. That's unnecessary. There just may not be people on at this time.

PLEASE HELP!!!

I'm not a 100 percent sure on this one, so you may wanna check my work before.  

Let's first start with the most restrictive restriction:

- There has to be a repeating pattern of 3 digits that is going to be repeated either by the second loop or the third loop, and stay in the first loop. 

The repeating pattern has 1600 possibilities because there are 8 possibilities for the hundreds digit (2-9), and 10 possibilities for the ones and tens digit (0-10). 

8 * 10 * 10 = 800

We have to multiply by 2, because the repeating pattern can either be in the second loop of 3 digits or the 2nd one. 

Therefore 800*2 = 1600 phone numbers

Problem: x/13 = 69

First step: Multiple 13 on both sides

x/13 * 13 = 69 * 13

x = 69*13

x = 897

Therefore the missing variable (x) is 897. 

We can assign the length and width variables, l and w respectively. 

The increase in each dimension would result in:

(l*1.2) * (w*1.2) = lw * 1.44

Since we know that lw is the original area, there is a 44 percent increase. 

We don't even need to consider the fact that the dimensions are 24 and 16!

Ako, I understand your process, but the last part of arithmetic is wrong. 

We know that the first term in the sequence in numbers between is 17/3 which's ceiling is 6, and (17/3)^2, which's floor is 32. 

The sequence is as follow: 

(6, 7, 8, ...., 30, 31, 32)

So we have to find the number of terms between 6 and 32 inclusive, which is 32 - 6 + 1 and that is 27, we have to do this because subtraction only includes 1 of the values not both of them. 

Or we could've done it by subtracting 5 from every term in the sequence then you would get:

(6-5, 7-5, 8-5, ...., 30-5, 31-5, 32-5)

(1, 2, 3, ..., 25, 26, 27)

Either way you can see that there are 27 terms, so the answer is 27 integers. 

We know that working 8 hours per day for 18 days results in 144 days. 

Since we are increasing the number of typists by 16/12 or 4/3 the number of hours they work per day must be inversely proportional. So we know that if there are 12 typists we need 4/3 times more hours than if we have 16 typists. Therefore we have to multiply by the number of hours that it takes 12 typists by 3/4 to solve. 

144 * 3/4 = 108

Now we know it takes 108 hours in total, and that means that 108/18 hours or 6 hours per day. 

Answer:

6 hours per day! :)

Also, note that instead of converting the number of hours per day to the total amount of hours, since the number of days they are working constant, we could've just kept it constant and ignore the number of hours they are working per day, so then it will be 8 * 3/4 which gives us 6 hours as previously predicted. 

50x + 1250 <= 12500

50x <= 11250

x <= 225

This means that each student has to raise at least 225 dollars!!

Enjoy!!

(Also, it would be great if someone told me to do the greater than/equal to sign because when I type it is <=)

Thank you!! That makes a lot of sense

Extend CE and extend AB, let us call the point of intersection \(L\)

 - Because of vertical angles angle \(LCB\) is 55 degrees.

 - Angle \(CBL\) is 180 - 120 = 60 degrees.

Since all angles in a triangle add up to 180 degrees, angle \(CLB\) = 65 degrees. Because of alternate interior angles, the supplement of CLB is congruent to CEF.

Therefore, CEF = 180 - 65 = \(\boxed{115}\) degrees.

That's what I initially thought but it was incorrect

thank u

Find sin(2x), cos(2x), and tan(2x) from the given information. sec(x) = 4, x in Quadrant IV

sin(2x) =

cos(2x) =

tan(2x) =

Hello Guest!

\(sin(2x)=2sin(x)cos(x)=2\frac{\pm \sqrt{sec^2(x)-1}}{sec(x)}\times\frac{1}{sec(4)}\\ sin(2x) =2\frac{-\sqrt{4^2-1}}{4}\times\frac{1}{4}\\\)

\(sin(2x)_{sec(x)=4}=-0.4841\)

\(cos(2x)=2cos^2(x)-1=2\times \frac{1}{sec^2(x)}-1\\ cos(2x)=2\times \frac{1}{4^2}-1\\\)

\(cos(2x)_{sec(x)=4}=-0.875\)

\(tan(2x)=\frac{2tan(x)}{1-tan^2(x)}=\frac{2\cdot (\pm\sqrt{sec^2(x)-1})}{1-(sec^2(x)-1)}\)

\(tan(2x)=\frac{2\times (-\sqrt{4^2-1})}{1-(4^2-1)}\)

\(tan(2x)_{sec(x)=4}=0.55328\)

  !

i cant faound it but thanks anyway