All Answers 358156 Answers

Very impressive Hayley.  

I didn't know you could answer questions that are this advanced !

The only thing I would suggest is that you show this is a minimum (unless you give a reason, it could be a maximum!)

The ceiling, I'm in a maths lesson #last day of school

You would think the four freaken questions I asked be answered by now😥😥😔😔😔😔😔😜😀😜😀😀😀😜

Sup?😜😀😜😀😜😀😀

Very cute LOL

I remember when I was the only mathematician and there were no questions that I could not answer.

I remember Christmas 2013 when the forum was so dead that I was writing 'plays' and playing all the characters myself. LOL.

Arr those were the days.... :)

Maybe I should see if I can dig some of those out   ://///

They have probably long since gone with the rest of the 'trash'    

Thanks for the trrip down memory lane SpawnofAngel       :D

I appreciate everyone who hasn't posted answers to my last three posts.😜😀😜😀😜😀😜

Life is a state of being so I don't think so lol😜😀😜😀😀😜😀

Sorry,, but when someone answers that post can you include the definition of cosh?😜

Both of the solutions given by Melody are correct.  Here is an alternative derivation:

These values for h and k can be put back into the earlier equations to get the corresponding radii.

.

ten to the power two hundred and thirty five.

five times ten to the power two hundred and thirty five.

2 is the gradient 

-1 is the y-intercept

The problem only wants the equation of a circle, it means just 1 equation. Your first equation is right but I'm not sure about your second equation answer.

solve using Cramer's rule

x-2y+3z=6

2x+y-z=-3

x+y+z=6

\(\begin{array}{rcrcrcr} 1\cdot x &-& 2\cdot y &+& 3\cdot z &=& 6 \\ 2\cdot x &+& 1\cdot y &-& 1\cdot z &=& -3 \\ 1\cdot x &+& 1\cdot y &+& 1\cdot z &=& 6 \\ \end{array}\\ \small{ \begin{array}{lcl} \\ \text{Determinant denominator} &=& \begin{vmatrix} 1&-2&3 \\ 2&1&-1 \\ 1&1&1 \\ \end{vmatrix}\\ \\ &=& 1\cdot 1\cdot 1 + 2\cdot 1\cdot 3 +1\cdot (-2)\cdot (-1) -1\cdot 1\cdot 3 -2\cdot (-2)\cdot 1 -1\cdot 1\cdot (-1) \\ &=& 1 + 6 + 2 -3 +4 +1 \\ &=& 11 \\ \end{array} }\)

\(\small{ \begin{array}{lcl} x &=& \dfrac{ \begin{vmatrix} 6&-2&3 \\ -3&1&-1 \\ 6&1&1 \\ \end{vmatrix} }{11}\\\\ &=&\dfrac{ 6\cdot 1\cdot 1 + (-3)\cdot 1\cdot 3 +6\cdot (-2)\cdot (-1) -6\cdot 1\cdot 3 -(-3)\cdot (-2)\cdot 1 -6\cdot 1\cdot (-1) } {11}\\ &=&\dfrac{ 6 -9 + 12 -18 -6 +6 } {11}\\ \mathbf{x} & \mathbf{=} & \mathbf{-\dfrac{9}{11}}\\ \end{array} }\)

\(\small{ \begin{array}{lcl} y &=& \dfrac{ \begin{vmatrix} 1&6&3 \\ 2&-3&-1 \\ 1&6&1 \\ \end{vmatrix} }{11}\\\\ &=&\dfrac{ 1\cdot (-3)\cdot 1 + 2\cdot 6\cdot 3 +1\cdot 6\cdot (-1) -1\cdot (-3)\cdot 3 -2\cdot 6\cdot 1 -1\cdot 6\cdot (-1) } {11}\\ &=&\dfrac{ -3 +36 -6 +9 -12 +6 } {11}\\ \mathbf{y} & \mathbf{=} & \mathbf{\dfrac{30}{11}}\\ \end{array} }\)

\(\small{ \begin{array}{lcl} z &=& \dfrac{ \begin{vmatrix} 1&-2&6 \\ 2&1&-3 \\ 1&1&6 \\ \end{vmatrix} }{11}\\\\ &=&\dfrac{ 1\cdot 1\cdot 6 + 2\cdot 1\cdot 6 +1\cdot (-2)\cdot (-3) -1\cdot 1\cdot 6 -2\cdot (-2)\cdot 6 -1\cdot 1\cdot (-3) } {11}\\ &=&\dfrac{ 6 +12 +6 -6 +24 + 3 } {11}\\ \mathbf{z} & \mathbf{=} & \mathbf{\dfrac{45}{11}}\\ \end{array} }\)

Check:

\(\begin{array}{rcrcrcr} 1\cdot ( -\dfrac{9}{11} ) &-& 2\cdot \dfrac{30}{11} &+& 3\cdot \dfrac{45}{11} &=& 6 \\ 2\cdot ( -\dfrac{9}{11} ) &+& 1\cdot \dfrac{30}{11} &-& 1\cdot \dfrac{45}{11} &=& -3 \\ 1\cdot ( -\dfrac{9}{11} ) &+& 1\cdot \dfrac{30}{11} &+& 1\cdot \dfrac{45}{11} &=& 6 \\ \end{array}\\\)

To find the inverse of a function switch x and y and then solve for y

Example:  \(y=3x+2\)

Switch x and y

\(x=3y+2\)

subtract 2 from both sides

\(x-2=3y\)

divide 3 from both sides

\((x-2)/3=y\)

rewrie with y on the left

\(y=(x-2)/3\)

The inverse of \(y=3x+2\) is \(y=(x-2)/3\)

-2(3)+3(-2)-6=

-6-6-6=

-12-6= -18

\(-2x+3y-6\); \(x=3\), \(y=-2\)

\(-2*3+3*(-2)-6\)

\(-6+3*(-2)-6\)

\(-6+-6-6\)

\(-6-6-6\)

\(-12-6\)

\(-18\)

what is *****-3x+(-6)-(-8)-6********

Solve for m:
5040000 = 1800 m

5040000 = 1800 m is equivalent to 1800 m = 5040000:
1800 m = 5040000

Divide both sides of 1800 m = 5040000 by 1800:
(1800 m)/1800 = 5040000/1800

1800/1800 = 1:
m = 5040000/1800

The gcd of 5040000 and 1800 is 1800, so 5040000/1800 = (1800×2800)/(1800×1) = 1800/1800×2800 = 2800:
Answer: | m = 2800
 

You can view your graph here:https://www.desmos.com/calculator

\(-3x+-6--8-6\)

\(-3x-6--8-6\)

\(-3x-14-6\)

\(-3x-20\)

Solve the following system: WHOSE IDEA WAS IT TO USE CRAMER'S RULE?
{x-2 y+3 z = 6
2 x+y-z = -3
x+y+z = 6
Express the system in matrix form:
(1 | -2 | 3
2 | 1 | -1
1 | 1 | 1)(x
y
z) = (6
-3
6)
Solve the system with Cramer's rule:
x = 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1/1 | -2 | 3
2 | 1 | -1
1 | 1 | 1 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/1 | -2 | 3
2 | 1 | -1
1 | 1 | 1 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/1 | -2 | 3
2 | 1 | -1
1 | 1 | 1
Use cofactor expansion about the 1^st row. 1 | -2 | 3
2 | 1 | -1
1 | 1 | 1  =  1 | -1
1 | 1+2 2 | -1
1 | 1+3 2 | 1
1 | 1:
x = 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1/1 | -1
1 | 1+2 2 | -1
1 | 1+3 2 | 1
1 | 1 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/1 | -1
1 | 1+2 2 | -1
1 | 1+3 2 | 1
1 | 1 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/1 | -1
1 | 1+2 2 | -1
1 | 1+3 2 | 1
1 | 1
2 | -1
1 | 1  =  2×1-(-1)  =  3:
x = 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1/(1 | -1
1 | 1+2×3+3 2 | 1
1 | 1) and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/(1 | -1
1 | 1+2×3+3 2 | 1
1 | 1) and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/(1 | -1
1 | 1+2×3+3 2 | 1
1 | 1)
2 | 1
1 | 1  =  2×1-1×1  =  1:
x = 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1/(1 | -1
1 | 1+2×3+3) and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/(1 | -1
1 | 1+2×3+3) and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/(1 | -1
1 | 1+2×3+3)
1 | -1
1 | 1  =  1×1-(-1)  =  2:
x = 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1/(2+2×3+3) and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/(2+2×3+3) and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/(2+2×3+3)
2×3 = 6:
x = 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1/(2+6+3) and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/(2+6+3) and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/(2+6+3)
2+6+3 = 11:
x = 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
Use cofactor expansion about the 1^st row. 6 | -2 | 3
-3 | 1 | -1
6 | 1 | 1  =  6 1 | -1
1 | 1+3 -3 | 1
6 | 1+2 -3 | -1
6 | 1:
x = 6 1 | -1
1 | 1+3 -3 | 1
6 | 1+2 -3 | -1
6 | 1/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
1 | -1
1 | 1  =  1×1-(-1)  =  2:
x = (3 -3 | 1
6 | 1+2 -3 | -1
6 | 1+2 6)/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
-3 | -1
6 | 1  =  -(-6)-3×1  =  3:
x = (3 -3 | 1
6 | 1+3 2+6 2)/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
-3 | 1
6 | 1  =  -3×1-1×6  =  -9:
x = (6×2+2×3+3×-9)/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
6×2 = 12:
x = (12+2×3-9×3)/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
2×3 = 6:
x = (12+6-9×3)/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
3 (-9) = -27:
x = (12+6+-27)/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
12+6-27 = -9:
x = -9/11 and y = 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
Use cofactor expansion about the 1^st row. 1 | 6 | 3
2 | -3 | -1
1 | 6 | 1  =  -3 | -1
6 | 1-6 2 | -1
1 | 1+3 2 | -3
1 | 6:
x = (-9)/11 and y = -3 | -1
6 | 1-6 2 | -1
1 | 1+3 2 | -3
1 | 6/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
2 | -1
1 | 1  =  2×1-(-1)  =  3:
x = (-9)/11 and y = (-3 | -1
6 | 1-63+3 2 | -3
1 | 6)/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
2 | -3
1 | 6  =  2×6--3×1  =  15:
x = (-9)/11 and y = (-3 | -1
6 | 1-6×3+3×15)/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
-3 | -1
6 | 1  =  -(-6)-3×1  =  3:
x = (-9)/11 and y = (3-6×3+3×15)/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
-6×3 = -18:
x = (-9)/11 and y = (3+-18+3×15)/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
3×15 = 45:
x = (-9)/11 and y = (3-18+45)/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
3-18+45 = 30:
x = (-9)/11 and y = 30/11 and z = 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6/11
Use cofactor expansion about the 1^st row. 1 | -2 | 6
2 | 1 | -3
1 | 1 | 6  =  1 | -3
1 | 6+2 2 | -3
1 | 6+6 2 | 1
1 | 1:
x = (-9)/11 and y = 30/11 and z = 1 | -3
1 | 6+2 2 | -3
1 | 6+6 2 | 1
1 | 1/11
2 | -3
1 | 6  =  2×6--3×1  =  15:
x = (-9)/11 and y = 30/11 and z = (1 | -3
1 | 6+2×15+6 2 | 1
1 | 1)/11
2 | 1
1 | 1  =  2×1-1×1  =  1:
x = (-9)/11 and y = 30/11 and z = (1 | -3
1 | 6+2×15+6)/11
1 | -3
1 | 6  =  1×6--3×1  =  9:
x = (-9)/11 and y = 30/11 and z = (9+2×15+6)/11
2×15 = 30:
x = (-9)/11 and y = 30/11 and z = (9+30+6)/11
9+30+6 = 45:
Answer: | x = (-9)/11   and y = 30/11   and z = 45/11
 

Es aquel en el cual la vista observa a un objeto hacia abajo de la recta horizontal

Let b = number of miles Jillian bikes in the evening for one day; Let m = total number of miles; Let d = number of total days; Let r = number of miles Jillian runs in the morning for one day

\(b=(m-dr)/d\)

\(b=(30-3*5)/5\)

\(b=(30-15)/5\)

\(b=15/5\)

\(b=3\)

Jillian bikes for a total of 3 miles each evening