All Answers 358187 Answers

5x+2x-9=40

combine like terms:

7x-9=40

add 9 on both sides:

7x=49

divide 7 on both sides:

x=7

log(a)*log(b)

if there and 9 horses in all 10 stables then there are 90 horses i think

~Laura

The two given distances are legs of a right triangle.....the distance that the plane is from the second station is just the hypotenuse of a right triangle......and this distance =

sqrt ( 115^2 + 136^2 )   =   about  178.1 miles

The angle we are interesed in is the one with the plane at its vertex.......this is given by

arctan (136/115) = about 49.78°

Since bearing is measured clockwise from a north-south line.......the bearing of the second station from the airplane =

[270 + 49.78]°  =  319.78°

Perhaps they sold only  200 tickets?

d = day tix

200-d = night tix

d (800) + (200-d)550 = 153900

800d + 110000 - d550 =153900

d250 = 43900

d= 175.6     you can't sell a PARTIAL ticket so....

                     Alan is correct......even with this assumption of 200 tix, the numbers still don't work

= 5a - b/12

To find the ANGLE from the airplane to the second radio station:

tan x  =  136/115  (opposite/adjacent)     

x=  49.78 degrees    FROM DUE WEST    180 - 49.78 = 130 .22 degrees is the COMPASS BEARING from the plane

Two ways to find the DISTANCE  :  1: Pythagorean Theorem

or   Sin 49.78 = opposite/hypotenuse = 136/hypotenuse      

  136/ (sin 49.78) = hypotenuse      (the DISTANCE)  = 178.11 miles

Find the smallest positive integer that satisfies the system of congruences

a)

n congruency sign 2 (mod ll)

n congruency sign 3 (mod 17)

\(\begin{array}{rcll} n &\equiv& {\color{red}2} \pmod {{\color{green}11}} \\ n &\equiv& {\color{red}3} \pmod {{\color{green}17}} \\ \text{Let } m &=& 11\cdot 17 = 187 \\\\ \end{array}\)

Because 11 and 17 are relatively prim ( gcd(11,17) = 1! ) we can go on:

\(\begin{array}{rcll} n &=& {\color{red}2} \cdot {\color{green}17} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ {\color{green}17}^{\varphi({\color{green}11})-1} \pmod {{\color{green}11}} ] }_{=\text{modulo inverse 17 mod 11} } }_{=17^{10-1} \mod {11} }}_{=17^{9} \mod {11}}}_{=2} + {\color{red}3} \cdot {\color{green}11} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ {\color{green}11}^{\varphi({\color{green}17})-1} \pmod {{\color{green}17}} ] }_{=\text{modulo inverse 11 mod 17} } }_{=11^{16-1} \mod {17} }}_{=11^{15} \mod {17}}}_{=14}\\\\ n &=& {\color{red}2} \cdot {\color{green}17} \cdot [ 2] + {\color{red}3} \cdot {\color{green}11} \cdot [14] \\ n &=& 68 + 462 \\ n &=& 530 \\\\ && n \pmod {m}\\ &=& 530 \pmod {187} \\ &=& 156 \\\\ n &=& 156 + k\cdot 187 \qquad k \in Z\\\\ \mathbf{n_{min}} & \mathbf{=}& \mathbf{156} \end{array}\)

b)

n congruency sign 1 (mod 7)

n congruency sign 7 (mod 13)

n congruency sign 13 (mod 20)

\(\begin{array}{rcll} n &\equiv& {\color{red}1} \pmod {{\color{green}7}} \\ n &\equiv& {\color{red}7} \pmod {{\color{green}13}} \\ n &\equiv& {\color{red}13} \pmod {{\color{green}20}} \\ \text{Let } m &=& 7\cdot 13\cdot 20 = 1820 \\ \end{array} \)

Because 7 and 13 and 20 are relatively prim  we can go on:

\(\small{ \begin{array}{l} n = {\color{red}1} \cdot {\color{green}13\cdot 20} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}13\cdot 20) }^{\varphi({\color{green}7}) -1 } \pmod {{\color{green}7}} ] }_{=\text{modulo inverse }(13\cdot 20) mod 7 } }_{=(13\cdot 20)^{6-1} \mod {7}} }_{=(13\cdot 20)^{5} \mod {7}} }_{=(260\pmod{7})^{5} \mod {7}} }_{=(1)^{5} \mod {7}} }_{=1} + {\color{red}7} \cdot {\color{green}7\cdot 20} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}7\cdot 20) }^{\varphi({\color{green}13}) -1} \pmod {{\color{green}13}} ] }_{=\text{modulo inverse } (7\cdot 20) mod 13 } }_{=(7\cdot 20)^{12-1} \mod {13}} }_{=(7\cdot 20)^{11} \mod {13}} }_{=(140\pmod{13})^{11} \mod {13}} }_{=(10)^{11} \mod {13}} }_{=4} + {\color{red}13} \cdot {\color{green}7\cdot 13} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ { (\color{green}7\cdot 13) }^{\varphi({\color{green}20}) -1 } \pmod {{\color{green}20}} ] }_{=\text{modulo inverse } (7\cdot 13) mod 20 } }_{=(7\cdot 13)^{8-1} \mod {20}} }_{=(7\cdot 13)^{7} \mod {20}} }_{=(91\pmod{20})^{7} \mod {20}} }_{=(11)^{7} \mod {20}} }_{=11}\\\\ n = {\color{red}1} \cdot {\color{green}13\cdot 20} \cdot [1] + {\color{red}7} \cdot {\color{green}7\cdot 20} \cdot [4] + {\color{red}13} \cdot {\color{green}7\cdot 13} \cdot [11] \\ n = 260+ 3920 + 13013 \\ n = 17193 \\\\ n \pmod {m}\\ = 17193 \pmod {1820} \\ = 813 \\\\ n = 813 + k\cdot 1820 \qquad k \in Z\\\\ \mathbf{n_{min}} \mathbf{=} \mathbf{813} \end{array} }\)

\(\begin{array}{l} \text{In number theory, }\\ \text{Euler's theorem (also known as the Fermat–Euler theorem or Euler's totient theorem)}\\ \text{states that if } \mathbf{n} \text{ and } \mathbf{a} \text{ are coprime positive integers, then }\\ \hline a^{\varphi (n)} \equiv 1 \pmod{n} \end{array}\)

The numbers don't stack up here.  Even if they only sold night tickets, 2000 of them would come to much more than153900!!

Well he has a gun wound to his head so I am guessing he died from that. All those other irrelevant details are just irrelevant. The fact you tell us not to google it shows your lackof originality.

The question is actually 10 horses in 9 stables and this is how you do it:

[T][E][N][H][O][R][S][E][S]

If this is meant to be 3^(2[x+1]) - 8*3^(x+1) = 9 then:

.

I just realised that your brackets are not paired so you probably meant a different problem   

3^(2(x+1)-8(3^(x+1))=9

\(3^{[2(x+1)-8*3^{(x+1)}]}=9\\ 3^{[2(x+1)-8*3^{(x+1)}]}=3^2\\ 2(x+1)-8*3^{(x+1)}=2\\ 2[(x+1)-4*3^{(x+1)}]=2\\ (x+1)-4*3^{(x+1)}=1\\ x+1-4*3^{(x+1)}=1\\ x-4*3^{(x+1)}=0\\ x=4*3^{(x+1)}\\ \frac{x}{4}=3^{(x+1)}\\ log(\frac{x}{4})=log(3^{(x+1)})\\ log(x)-log(4)=(x+1)log(3)\\ log(x)-log(4)=xlog(3)+log(3)\\ log(x)-log(4)=log(3^x)+log(3)\\ log(x)-log(3^x)=log(3)+log(4)\\ log(\frac{x}{3^x})=log(12)\\ \frac{x}{3^x}=12\\\)

And that is about as far as I can go  :/

b)

n congruency sign 1 (mod 7)                   n=7p+1

n congruency sign 7 (mod 13)                 n=13q+7

n congruency sign 13 (mod 20)               n=20r+13

13,33,53,73,93,113,133,153,173,193,...............293

which is a multiple of 7

12,32,52,72,92, 112, ....... ...................................292

which is a multiple of 13

6,26,46,66,86,106,126,146,166,186,206,226, 246,266,286,

33+13*20n = 33+260n   will    satisfy    13(mod20) and  7(mod 13)

33,293,553,813,1073

33+260n   which of these will be 1 mod 7

32+260m must be a multiple of 7

32+260m must be a multple of 7

260m+32=7k

292, 552, 812,

812+1=813

813 = 1(mod 7)

813 =  13(mod 20)

813 =  7(mod 13)

The smallest number that meets those criterion is  813

There is probably a much better way to do it though ://

Here are a couple of examples where it does just that:

.

NO! 

Find the smallest positive integer that satisfies the system of congruences

a)

n congruency sign 2 (mod ll)      n= 11p+2

n= 2,13,24,35,46,57,68,79,90,101,112,123,134,145,156,
n congruency sign 3 (mod 17)    n=17q+3

n=3,20,37,54,71,88,105,122,139,156,173,190

Surface area = 4 π r²

Volume = 4/3 π r³

8x - 3y = 38

3x - 2y = -1

8x = 38 + 3y

3x = -1 + 2y

5x = 39 + y

2x = 40 - y

x = 20 - 0.5y

Substutute x with 20 - 0.5y because x = 20 - 0.5y

3x = -1 + 2y

3 × (20 - 0.5y) = -1 + 2y

60 - 1.5y = -1 + 2y              |    +1

60 - 1.5y + 1 = 2y

61 - 1.5y = 2y                      | +1.5 y

61 = 3.5y

y = 122/7

Substitute y with 122/7 because y  =122/7

x = 20 - 0.5y

x = 20 - 0.5 × 122/7

x = 20 - 61/7

x = (140-61)/7

x = 79/7

---------------------------------------------------------------

Solution

x = 79/7

y = 122/7

3 + 1 = 4 units → 128

1 unit → 128 ÷ 4 = 32 (Red Engines)

3 units → 32 × 3 = 96 (Blue Team)

Maybe he stole a bunch of bycicles from some native aericans way back when, and got shot for doing that and was covered in animal dung...

Well the chips have to be something other than like poato chips... right?

Like, maybe... Buffalo chips? Poker chips?