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(5x+-2)-(4x+14)

=(5x-2)   -1(4x+14)

=5x-2       -4x-14

=5x-4x -2-14

=1x-16

=x-16

$${{\mathtt{504}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = {\mathtt{7.958\: \!114\: \!415\: \!792\: \!783}}$$

that answer is approximate :)

I'll help with my thumb too :)

Thanks Alan and Heureka   

$${\frac{{{\mathtt{0.2}}}^{\left({\mathtt{1}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{2.5}}}}\right)}}{{\sqrt[{{\mathtt{3}}}]{{{\mathtt{125}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{0.3}}}^{{\mathtt{3}}}}}}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\frac{{\mathtt{\pi}}}{{\sqrt{{\mathtt{5}}}}}}\right)} = -{\mathtt{0.897\: \!779\: \!501\: \!829\: \!020\: \!9}}$$

try i am learning and awnswer is id not care

Thank you for the history lesson Heather  

I was wondering what the Cinco De Mayo (5th of May) memorial day was all about.     

You have repeated this question without stating that it is a repeat question.

PLEASE DO NOT DO THIS!!

Alan and i have both answered you and as a result other question will not be answered.

There is not time to answer all!

You are invited to repost - that is fine BUT  please follow the intstuctions that i have laid out here.

NOTE:

Equations that can only have integer solutions are call diophantine equations

As a general rule thes are much more difficult to solve than normal equations where all real solutions are acceptable.

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I have just seen a pop-up telling me that you have repeated this question without stating that it is a repeat question.

PLEASE DO NOT DO THIS!!

Alan and i have both answered you and as a result other question will not be answered.

You are invited to repost - that is fine BUT  please follow the intstuctions that i have laid out here.

Question:

 u(x,0) = e^{(-x^2)}  ,  u_t(x,0) = d/dx(e^{-x^2})

Solve  using d'Alembert's solution

u(x,t) = 1/2[f(x-at)+f(x+at)] + 1/2a $$\int_{x-at}^{x+at} \mathrm{g}{(s)}\,\mathrm{d}s$$  In this case a = 1.

$$u(x,0) = g(x) = e^{-x^2} \qquad
\boxed{
g(x-t) = e^{-(x-t)^2} \qquad
g(x+t) = e^{-(x+t)^2} }$$

$$u_t(x,0) = h(x) = \dfrac{d\left( e^{-x^2} \right)}{dx} \qquad
\boxed{
\int_{x-t}^{x+t} h(\xi) \, d\xi. = \int_{x-t}^{x+t} \dfrac{d\left( e^{-\xi^2} \right)}{d\xi} = \left[ e^{-\xi^2} \right]^{x+t}_\limits_{x-t}\\
}$$

$$u(x,t) = \frac{1}{2}\left[g(x-t) + g(x+t)\right] + \frac{1}{2} \int_{x-t}^{x+t} h(\xi) \, d\xi.\\\\
u(x,t) = \frac{1}{2}\left[ e^{-(x-t)^2} + e^{-(x+t)^2}\right]
+\frac{1}{2}\left[ e^{-\xi^2} \right]^{x+t}_\limits_{x-t}$$

$$u(x,t) = \frac{1}{2} \left[ e^{-(x-t)^2} + e^{-(x+t)^2}\right]
+\frac{1}{2} \left[ e^{-(x+t)^2} - e^{-(x-t)^2} \right]\\\\
u(x,t) = \frac{1}{2} \left[ e^{-(x+t)^2}\right]
+\frac{1}{2} \left[ e^{-(x+t)^2} } \right]\\\\
\boxed{ u(x,t) = e^{-(x+t)^2} }\\\\
u(x,0) = e^{-(x)^2}\\
u(x,1) = e^{-(x+1)^2}\\
u(x,10) = e^{-(x+10)^2}$$

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ok

spiders have 8 legs and 1 head               S

beatles have 6 legs and one head            B

apodous larvae have no legs and 1 head     L

S+B+L=17    and     8S+6B=58

Lets look at this     8S+6B=58      

number of spiders that there could be             1     2     3     4     5     6     7 

no. of spiders legs                                            8   16   24   32   40   48  56 

legs left for beatles                                         50   42  34   26     18   10  2  

Which of these is divisable by 6?                            42                   18

How many beatles might this be (divide by 6)         7                     3

SO there could be     2 spiders 7 beeetles and 8 Lavae

or  there could be     5 spiders  3 beetles and  9 lavae    

She has less lavae than spiders and beetles combined so 

SO there must be     2 spiders 7 beeetles and 8 Lavae

Expand the terms in parentheses;

x³ - 3x² + 3x -1 - x³ + 4x² -4x = 1 - 2x

Collect like terms

x² + x - 2 = 0

This factors nicely:

(x - 1)(x + 2) = 0

so x = 1 and x = -2 are the solutions

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idk solve it out

What's the question?!!!

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As all prime numbers except the number 2 are odd, any two (excluding the number 2) must sum to an even number.  Therefore to have two prime numbers sum to an odd number, one of them must be 2.  This would mean that the other number would be 287.  However, 287 is not a prime number, so no two prime numbers add up to 289.

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Welcome back Db

TRue

-4^2 = -1*4^2=-1*16=-16

and 

-16<-4

it is    $${{\mathtt{x}}}^{{\mathtt{y}}}$$

Hey MathsGod1 , take a look at the LaTex for this.  THEN you will know how to do greek letters :)

$$lambda\;is\;\lambda$$

it is a greek letter