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Let  t  be the number of hours, and let  v  be the car's velocity in kilometers per hour.

When  t = 3 ,  v = 53  .

When  t = 6 ,  v = 62 .

So between the hours of  3  and  6 , the car's velocity changes at a rate of

\(\frac{ 62-53}{6-3}\,=\,\frac93\,=\,3\)  kilometers per hour per hour

And...if the velocity of the car changes at a constant rate....

3  =  the rate of change between any two possible values of  v  and  t .

3  =  \(\frac{62-v}{6-t}\)

18 - 3t  =  62 - v

-44 - 3t  =  -v

v  =  3t + 44

Thank you, but I would prefer not to list every single combination and then count them all up, as that would take a long time. I would believe there is some sort of combinatorical way to solve this problem, I just don't know how.

well there are two ways you can do it the first way is to do the first one and the second is to do is the second one but instead of useing three ones use a two and a one on the one where it says 1+1+1+17 change it to 17+2+1.

y  =  x  + 4/x^2

Write as

y  = x  + 4x^(-2)

y ' =  1  - 8x^(-3)        

Set this  = 0

1 -  8x^(-3)  = 0  

1  =  8 / x^3  multiply both sides by x^3

x^3  = 8    ⇒    x  = 2

And when  x  = 2,  y  =  2 + 4 /(2)^2   =   2  + 1   = 3

Take the second deriviative

y''  =  24x^(-4)  =  24 / x^4

Putting 2 into this gives a positive.....so....this indicates that (2,3)  is a minimum

Here"s the graph : https://www.desmos.com/calculator/3xcrgvdoyl

This is "solvable"  by at least two methods that I know of OBT, but it can be a little cumbersome with matrixes that are larger than 3 x 3

Here's a website that will help you : http://matrix.reshish.com/matrixMethod.php

Click on the tab that says  "Inverse Matrix Method"  and choose the matrix dimension  [ this one is "4"]......hit "Set Matrix" and a matrix will come up....you can fill in the coefficients and then hit "Solve"

The solution to this system is

x^3  =  5/3

x^2 = 0

x = -14/3

d  = 5

Here's another way

sin  = y / r

x  = - √2

y =  ???

r  = 5

y  =   +√ [ r^2 - x^2 ]

y = + √ [ 5^2 - (-√2)^2 ]      

y = + √ [25 - 2]

y = √23

So 

sin  = √23 / 5

A'B'C' can be found as follows :

A'  = 

( A - center of dilation)*scale factor + center of dilation =

[ (-4, -1)  - (-4, -3) ] * 2  +  (-4, 3)  =

[ (0 ,2) ]*2  + (-4, -3)  =

(0,4) + (-4, -3)  =

(-4, 1)

B' and C' are found similarly

B' = 

[ (-4, -3) - (-4, -3) ]*2 + (-4, -3) =

[ (0, 0)] *2   + (-4,-3)  =

(-4, -3)

C' =

[  (-1, -3) - (-4, -3)] *2  + (-4, -3)  =

[ (3, 0)] *2 + (-4, -3)  =

(6, 0) + (-4, -3)  =

(2. -3)

Here are the pre and post image points  : https://www.desmos.com/calculator/2nixwm8xtj

cos(−θ)=−√2/5 sin. θ>0

(−θ) =cos^-1(-√2/5)

(θ) =106.43 Degrees

sin(106.43) =√23/5

4/3     

The two triangles are congruent, so:

PR = SU

3y - 2 =y + 4    Subtract y from both sides 

3y - 2 - y = 4    Add 2 to both sides

3y - y =4 + 2

2y = 6                Divide both sides by 2

y =6/2

y = 3 

Sub 3 for y in 3y - 2

3*3 - 2 =7 Ft - Length of side PR

Perimeter of PQR =4 + 6 + 7 =17 Feet

Since they are similar (or proportionate ) triangles:

30 is to 25    as  4x-4 is to 20     or:

30/25 =(4x-4) / 20

(20)(30/25) = 4x-4

24 = 4x -4

28 = 4x

x=7

nevermind i have my answer. mark it as answered someone!

I'm having a Great day Thanks for asking!

Maybe this will clear up some of the issues

Do exactly the same with this one as "Guest" showed you how to do the other one. Just change the numbers.

11x - 3 = 9x + 23    Isolate x to the LHS. Subtract 9x from both sides.

11x - 9x - 3 = 23     Add 3 to both sides

2x =23 + 3

2x = 26                     Divide both sides by 2

x =26/2

x = 13

What is the value of X?

\(\begin{array}{rcll} 11x-3 &=& 9x+23 \\ 2x &=& 26 \\ x &=& 13 \\ \end{array}\)

X and Y are natural numbers.

X = 1; Y = 1

X = 2; Y = 2

X = 3; Y = 3

X = 4; Y = 4

X = 5; Y = 3

X = 6; Y = 2

X = 7; Y = 1

X = 8; Y = 0

How do I express this with a formula?

\(\begin{array}{|rcll|} \hline Y = \min(X,8-X) \\ \hline \end{array}\)

finding angle b in question 1?

\(\begin{array}{|rcll|} \hline \sin(B) &=& \frac{3.6}{4.2} \times \sin(110^{\circ}) \\ \sin(B) &=& 0.85714285714 \times 0.93969262079 \\ \sin(B) &=& 0.80545081782 \\\\ B &=& \arcsin(0.80545081782) \\ B &=& 53.6538157268^{\circ} \\\\ \mathbf{B} &\mathbf{=}& \mathbf{53.7^{\circ}} \\\\ A &=& 180^{\circ}- 110^{\circ} - 53.7^{\circ} \\ \mathbf{A} &\mathbf{=}& \mathbf{16.3^{\circ}} \\\\ \hline \end{array}\)

Good luck on your test Jeff. :)

C= 572.29 + service charge = 579.96        sorry...I added finance charge to AVERAGE instead of final balance !

Thanks Killersteve27 

Your introduction site looks really good too :)

C incorrect