All Answers 358093 Answers

6-1=5

4+5=9

5/9

Y-intercept is -7

do 4*(1+3^2)and 4*(1+3)^2 have the same answer

Why don't evaluate them separately and find out like this:

4*(1+3^2)=4* (1+9)=4*10=40

4*(1+3)^2=4*4^2=4*16 =64

Are they the same answer?

well said my friend  👏 👏

Bored?! Try hitting your books. That should wake you up pronto.

THAT NUMBER....TOO LARGE....HOLLY?? WHO 

IS HOLLY

HOLLY SOUNDS LIKE A BLAST NOT GONNA LIE 😍 😍👌

It can be solved through graphing and also through iteration and interpolation. Using the latter, it readily gives x=4

1. a snake

2. a hammer to make a hole in the wall

3. me

joke's on you I used bing 

Hmmmm     I know the answer is x=4    but I am drawing a blank on how to get there !    Anyone else?

please tell annabelle to stop it we didn't need this much punch

that girl is crazy sometimes

still love you tho annabelle 😍 😍

WOW!!! that is a lot of fruit 

I sure do hope that is it grapes and maybe apples

raise your hand if you want grapes and apples  ✋  ✋ ✋

I'm here!!! 😍 😍 🙌

welcome friend!!

x=wt of small box     y = large box wt

First delivery 

    5x  + 2y = 96      solve for y = (96-5x)/2

Second Delivery

   8y + 3x = 163       Substitute for y

   8 ( ( 96-5x)/2 ) + 3x = 163       and solve for x

    384 - 20x +3x = 163

17x=221

x=13   =   small box wt   kg               y= (96-5x)/2 = large box wt = 15.5 kg

i ont give  no @#$%!

give me something to do. anything.

In school we just started with this new topic, and I knew that we have this general formula for a second degree polynomial, so I thought that here may be the same for third degree polynomial. The links was very helpfull for me when I had to figure this topic out, thank you - I understand the topic now!

two apples

Solution:

\(R = \dfrac{110V}{1.6A} = 68.75 {~ohms} \\ \rho_{20} = 55.0*10^{-9} \Omega \cdot M \\ \alpha = \text {Temperature coefficient for tungsten} = 0.0045K^{-1} (K^{-1} \equiv C^{-1})\\ \rho_{1500}= (\rho_{20})*[1+(\alpha\Delta T)]\\ \rho_{1500}= (55.0*10^{-9})*(1+0.0045 C^{-1})*(1500-20)C =8.287125*10^{-5} \text {M} \cdot \Omega \\ A_c = (\pi * r^2) = \pi * (0.073*10^{-6})^2 = 1.67415473*10^{-8} M^2 \text {(Area cross-section)} \\\\ \)

\(R= \rho *(L*A) \rightarrow \text {Solve for length L. } \\ L = \dfrac {RA} {(\rho_{20}*[1 + \alpha * \Delta T])} \\ \hspace {10 mm} \\ \dfrac {(68.75)*(16.7415473*10^{-9})} {(55.0*10^{-9})*(1 + [0.0045 *1480])} =2.73198 ~\text {Meters} \\ \)

I want an answer smart a*s guest not another hint. 

There actually is a general solution for a third degree polynomal, but - it's messy ......you can read about it here :

http://www.math.vanderbilt.edu/~schectex/courses/cubic/

Here's another method using a "depressed" cubic: 

http://www.sosmath.com/algebra/factor/fac11/fac11.html

However.......there may be easier ways of solving certain 3rd degree polynomials -  for instance,  if they can be factored

Also......certain Calculus techniques will provide approximate solutions.....

Do you have a particular problem in mind???

Lel what?