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Find the sum of all positive rational numbers that are less than 5, and that have a denominator of 30 when written in lowest terms.

A question for you my friend.

Which of these 5  will be included in the sum?  There might be more than 1 correct answer

   \(1) \frac{1}{30}\\ 2) \frac{5}{30}\\ 3) \frac{30}{40}\\ 4) \frac{30}{30}\\ 5) 3\frac{7}{30}\)

Answer this for me and then someone (maybe me) will help you further.  (If you need more help)

Please no one answer over me.  I am trying to facilitate learning and interaction.

The chioces are 

A. logm​(3)

B. log3​(m)

C. -logm​(3)

D. -log3​(m)

The chioces are 

A. log_m​(3)

B. log₃​(m)

C. -log_m​(3)

D. -log₃​(m)

I'm just playing.  Like EP said, what are the choices?

If   

\(y=(log_3(m))^{-1}\\ then\\ \frac{1}{y}=log_3(m)\\ 3^{\frac{1}{y}}=m\\ m^y=3\\ \)

I think that is supposed to be   1/log₃ (m)     

Here is an example:

domain   x = [0,3]

range       y = [0,20]

Yes it is a function.....each value of x has only one corresponding y value

1/log3(m)=2.09590327 meters

What are the choices?

I'd be interested in seeing this one done too   

Oblique asymptotes are a mystery to me.

At the beginning $300 is invested.

If it is invested using simple interest then after 35 years it will grow to about $610    (I read that off the graph)    [that is $300 + $310 in interest]

If it is invested using scompound interest then after 35 years it will grow to about $870           [that is $300 + $570 in interest]

870- 610 =  $260               (this is also the same as 570-310)

So the compound interest investment grows $260 more than the simple interest one.

You can do the second one yourself.

1- 750 shares x $11.06 per share =$8,295 + $12 commission =$8,307.00 - Cost of buying the 750 shares.

2- 750 shares x $10.24 per share =$7,680 - $12 commission =$7,668.00 - What she got for selling them.

3 - $8,307.00 - $7,668.00 =$639.00 her LOSS for the year.

4 - 1 - [$7,668 / $8,295] =7.56% - what she LOST as a percentage for the year.

\(a)~\text{any positive integer power of 3:}~3, 9, 27, 81, \text{ etc.}\\ b)~\text{any negative integer power of 3:}~\dfrac 1 3,\dfrac 1 9,\dfrac{1}{27},\dfrac{1}{81}\\ c)~\text{any rational power of 3:}\sqrt{3},3\sqrt{3},\sqrt[3]{3}\text{ etc.}\)

\(2y-2a=6x\\ y = 3x+a\)

\(y+1=(a+6)x\\ y = (a+6)x-1\)

\(\text{The two lines are parallel so they have the same slope}\\ 3=(a+6)\\ a=-3\)

\(f(x)=3(x-2)^2-5\\ \text{This is the function}\\ y = x^2\\ \text{that has been shifted down 5 units, shifted to the right 2 units, and scaled by 3}\)

\(\text{The domain is all real numbers}\\ \text{The range is }[-5,\infty)\\ f(x) \text{ is a functional relation as each element in the domain maps to only a single element in the range}\)

\(\text{do you mean}\\ \sqrt{2^{i7\pi/4}}\)

\(-2-2i = 2\sqrt{2} e^{i5\pi/4}\\ z^3 = 2\sqrt{2} e^{i5\pi/4}\\ z = \sqrt{2} \exp\left(i\dfrac{\frac{5\pi}{4}+2k\pi}{3}\right), ~k=0,1,2\\ z =\sqrt{2}\exp\left(i\dfrac{5\pi}{12}\right),~ \sqrt{2}\exp\left(i\dfrac{13\pi}{12}\right),~\sqrt{2}\exp\left(i\dfrac{7\pi}{4}\right)\\ \text{and of these }\\ z = \sqrt{2}\exp\left(i\dfrac{7\pi}{4}\right) \text{ lies in the fourth quadrant}\)

Oh thanks a lot! Why didn't I think of that? ( time for bed... 😴 I think my brain malfunctioned...)

How many three-digit whole numbers have no 9's and no 7's as digit 

Not no 8

\(\text{There are 7 valid digits for the first number, and 8 valid digits for numbers 2 and 3}\\ \text{Thus there are }\\ n = 7\cdot 8^2 = 448 \text{ 3 digit numbers with no 7's or 9's}\)

You bsically have 8 digits that can go into each of the three positions (seven in the first one....no zero)

1-6 and 8    x  8   x  8    = 448 possibilities

the answer is take the hint I gave you and do your own #$)(*# work

3x + 2y =8        y = 4x-7    SUb second equation in to the first...then solve for x

3x + 2 (4x-7) = 8

3x +8x -14 = 8

11x = 22

x = 2                        Use this value of x to sub back in to one of the original equations to find y

y = 4x-7

y = 4(2) - 7

y = 1                         so the solution point is  (2,1)