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At first, I needed help with this confusing math problem about limits, but then I looked at it differently and found the solutions I needed to continue.

1. \(\displaystyle \lim_{x \to 2^{-}}g(x) = \) 4

2. \(\displaystyle \lim_{x \to 2^{+}}g(x) =\) -2

3. \(\displaystyle \lim_{x \to 2}g(x) =\) DNE

4. \(\displaystyle \lim_{x \to 0}g(x) = \) DNE

5. \(\displaystyle g(2) =\) 3

this problem is so frustrating

I know that te upper bound is when you use random numbers and then use synthetic divison and when the bttom part is all positive then thats the answer.. But what numbers do i exatl choose, do I just pick all number from 1-100 idk... I feel like that would take a while 

Thanks! That was really helpful

Thank you for helping me and writing out the math too!

b) Half of the numbers have an odd tens digit, so there are \(136 \div 2 = \color{brown}\boxed{68}\)

a) If the ones digit is a 5, there are 8 choices for the hundreds digit (can't be 0 or 5) and 8 choices for the tens digit. So there are \(8 \times 8 = 64\).

    If the ones digit is a 0, there are 9 choices for the hundreds digit and 8 for the tens digit which makes for \(9 \times 8 = 72\) choices

So, there are \(64 + 72 = \color{brown}\boxed{136}\) numbers 

Evaluate the expression.

Hello Keihaku!

\((\frac{16}{9})^{\frac{2}{3}}\cdot (\frac{4}{3})^{\frac{5}{3}}\\ =\sqrt[3]{\frac{4^{2\cdot 2}}{3^{2\cdot 2}}\cdot \frac{4^5}{3^5}}\\ =\frac{4^{(4+5)/3}}{3^{(4+5)/3}}\\ =\frac{4^3}{3^3}\\ \color{blue}=\frac{64}{27}=3.\overline{370}\)

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a):  

105 , 120 , 125 , 130 , 135 , 140 , 145 , 150 , 160 , 165 , 170 , 175 , 180 , 185 , 190 , 195 , 205 , 210 , 215 , 230 , 235 , 240 , 245 , 250 , 260 , 265 , 270 , 275 , 280 , 285 , 290 , 295 , 305 , 310 , 315 , 320 , 325 , 340 , 345 , 350 , 360 , 365 , 370 , 375 , 380 , 385 , 390 , 395 , 405 , 410 , 415 , 420 , 425 , 430 , 435 , 450 , 460 , 465 , 470 , 475 , 480 , 485 , 490 , 495 , 510 , 520 , 530 , 540 , 560 , 570 , 580 , 590 , 605 , 610 , 615 , 620 , 625 , 630 , 635 , 640 , 645 , 650 , 670 , 675 , 680 , 685 , 690 , 695 , 705 , 710 , 715 , 720 , 725 , 730 , 735 , 740 , 745 , 750 , 760 , 765 , 780 , 785 , 790 , 795 , 805 , 810 , 815 , 820 , 825 , 830 , 835 , 840 , 845 , 850 , 860 , 865 , 870 , 875 , 890 , 895 , 905 , 910 , 915 , 920 , 925 , 930 , 935 , 940 , 945 , 950 , 960 , 965 , 970 , 975 , 980 , 985 , Total =  136 such 3-digit numbers.

b):

130 , 135 , 150 , 170 , 175 , 190 , 195 , 210 , 215 , 230 , 235 , 250 , 270 , 275 , 290 , 295 , 310 , 315 , 350 , 370 , 375 , 390 , 395 , 410 , 415 , 430 , 435 , 450 , 470 , 475 , 490 , 495 , 510 , 530 , 570 , 590 , 610 , 615 , 630 , 635 , 650 , 670 , 675 , 690 , 695 , 710 , 715 , 730 , 735 , 750 , 790 , 795 , 810 , 815 , 830 , 835 , 850 , 870 , 875 , 890 , 895 , 910 , 915 , 930 , 935 , 950 , 970 , 975 , Total =  68 such 3-digit numbers

What is the radius of its orbit in meters?

Hello Guest!

\( F_{\rm{zp}} = m \cdot {\frac{v^2}{r}}\\ r = m \cdot {\frac{v^2}{F_{\rm{zp}}}}\\ r=1\ 500kg\cdot \dfrac{4600^2m^2}{150N\cdot s^2}\cdot\dfrac{N\cdot s^2}{kg\cdot m}\\ \color{blue }r=211\ 600\ 000m\)

The radius of its orbit is 211 600 000m.

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\(Ompute\ (\frac{8}{125})^{-\frac{1}{3}}\)

Hello Keihaku!

\({\color{blue}(\dfrac{8}{125})^{-\frac{1}{3}}}=\dfrac{1}{\sqrt[3]{\dfrac{8}{125}}}=\dfrac{1}{\dfrac{2}{5}}=\dfrac{5}{2}\color{blue}=2.5\)

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The smallest number  > 3456==3504

The system of quadratic equations     y = 3x^2 - 5x,      y = 2x^2 - x - c
c is a real number.

Hello Guest!

\( y = 3x^2 - 5x = 2x^2 - x - c\ |\ Equate\ equation\ to\ zero.\ |\color{blue} \ -(2x^2-x-c)\\ x^2-4x+c =0\ |\ p-q\ formula:\color{blue}x=-\frac{p}{2}\pm\sqrt{(\frac{p}{2})^2-q}\\ x=\frac{4}{2}\pm\sqrt{(\frac{4}{2})^2-c}\\\)

(a)

The system have exactly one real solution: 

\((\frac{4}{2})^2-c=0\\ c=(\frac{4}{2})^2\\ \color{blue}c=4\)

(b)

The system will have more than one real solution:

\((\frac{4}{2})^2-c>0 \)

- \(\color{blue}\infty\) < c < 4

(c)

The system will have no real solutions:

\((\frac{4}{2})^2-c<0\\ \color{blue}c>4 \)

  !

thanks

Let the speed of the current ==w

20 / (30 - w)  +   20 / (30  +  w)==1.5, solve for w

w == 10 mph - the speed of the current.

Are you the person that has posted this question 3 - 4 times ?? We told you under your your fiirst post that the sum of the squares CANNOT be 4 !! Why? Because:

2 ==F / [1 - R].................(1)

4 ==F^2 / [1 - R^2]........(2), solve for F, R

You get:

F ==2 - this is the first term

R==0 - this is the common ratio.

So: what does the above mean? It means that your infinte series has ONLY 1 term!!!!!. Which is 2, because the sum of the first term==2  and 2^2 ==4 !!!. Therefore, your question does not make sense the way it is written. The 4 in your question should be 3. Wherever you got it from, bring it to the attention of your Math Teacher.

An irrational number is a number that can not be expressed as the ratio of 2 integers. 

Examples include \(\pi\) and \(\sqrt {15}\).

\((\sqrt{3})^3 = 3 \sqrt 3 \)

\((\sqrt{3 \sqrt 3})^4 = ({3 \sqrt 3})^2 = \color{brown}\boxed{27}\)

Your sequence could be:

1/3, 4/5, 9/7, 16/9, 25/11, 36/13, 49/15, 64/17, 81/19, 100/21, 121/23, 144/25, 169/27.........etc.

The "closed form" of the sequence could be:

a(n) = n^2/(2 n + 1) (for any term)

Original price   x

Price in 2001   0.8*x

Price in 2002  0.8*0.8*x

Price in 2003    f*0.8*0.8*x = x  so f = 1/(0.8*0.8) = 1.5625  

I'll let you work out the percentage increase from here.

For equation \(ax^2 + bx + c = 0\)  to have repeated roots we must have \(b^2 - 4ac = 0\)

Calculate this for each quadratic in turn.

Add the two equations together.  Divide both side of the result by 4 to get x+y+z, then divide by 3 to get the average.

Hello asinus,

I do appreciate your response and explanation. I'm not sure how I saw this, because I input the measurments in an app, and got an error message. AAAHHH, wait, something just occured to me now while typing here...OBVIOUSLY the long side cannot be greater or equal than the two shorter sides combined, because if it was at least equal, we would have two straight parrallel lines....NOT exceeding, in this case, 200. COMMON SENSE ....

Gosh, forgive my stupidity please, wasting your time on something like this. I do appreciate your time. 

 How many pounds do Abby and Damon weigh together?

Hello weredumbmaster!

a + b = 260      a = 260 - b     

b + c = 245

c + d = 270      d = 270 - c

                  a + d = 260 + 270 - (b + c)

                  a + d = 260 + 270 - 245

                  a + d = 285

285 pounds do Abby and Damon weigh together.

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