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\(b^x=y\) is the exponential form.

\(log_by=x\) is the log form.

The exponential form is \(12^2=144\).

You likely switched the X and Y when you wrote the log formula.

So since the y is 144,

you were supposed to have \(log_{12}\boxed{144}=2\)

The 144 is the part you messed up, as you switched the x and y.

Substitute second equation into the first one

y-(4y-2)=14

-3y-2=14

-3y=16

y=-16/3

Now substitute back

x=-70/3

This is a simple algebra problem. You set X to set and get the value for X. You do the same thing for Y. 

Once you get these vales, just plug them into your problem. 

1500 = 1000 * e^(.03t)       divide both sides by 1000

1.5 = e^(.03t)       take the Ln of both sides

Ln 1.5  =  Ln e^(.03t)      and we can write

Ln 1.5  = (.03t) Ln e        [ Ln e  = 1....so we can forget this  ]

Ln 1.5  = .03t      divide both sides by .03

[ ln 1.5 ] / .03  = t ≈  13.52  years

We can equate slopes to find p

[ - 14 -- 5]       [ -17 - - 5]

________  =__________     simplify

    p -  2           (p + 2) - 2

[ -9]             [ - 12 ]

____   =    _______               cross- multiply

p - 2               p

-9p  =  -12 [ p - 2]

-9p  =  -12p + 24         add 12p to both sides

3p  =  24     divide both sides by 3

p = 8

So...the slope   =  -12 / 8   =   -3 / 2

So  since the points ( 2, - 5)  and ( 13, q)   are on the same line....we have....

[q - - 5 ]          -3

_______  =   ___ 

 13  - 2           2

[ q + 5 ]         -3

______  =      ___           cross-multiply

   11                2

2 [ q + 5 ]  =  -3 * 11

2q + 10  = -33

2q  =  -43

q = -43 / 2

So p + q  =   8  - 43/2  =    [ 16 - 43 ] / 2  =   -27/2  =   -13.5

Hint: They all have the same slope.

Non-Trig Solution:

Draw an octagon...like the diagram as shown below. Since the perimeter of the octagon is 32 cm, each of its side length are  32 / 8 = 4 centimeters.

Now, draw a square around it...forming four 45-45-90 degree triangles. The side length opposite to the 45 degree angles is \(x\sqrt{2}=4, x=\frac{4}{\sqrt{2}}=2\sqrt{2}\).

Find the area of each triangle, and multiply that by 4, yielding...\(4*4=16.\)

The area of the square on the other hand is \((4\sqrt{2}+4)^2=48+32\sqrt{2}.\)

Finally, (the area of the square) - (area of the four triangles = (area of the octagon).

Thus, we have \(48+32\sqrt{2}-16=\boxed{32+32\sqrt{2}}.\)

The reflection of  ( x, y)   about the line y = x   is  ( y, x)

So   (6, -5)  ⇒ (-5, 6)

Yes, it is simple common sense. If you double the amount you borrow, then the payment will also double, provided that the term and the interest rate remain the same for both loans. Here are a couple of examples:
N=5*12; R=0.06/12;PV=5000; PMT=PV*R*((1 + R)^N) / ((1 + R)^(N) - 1)
;N=5*12; R=0.06/12;PV=10000; PMT=PV*R*((1 + R)^N) / ((1 + R)^(N) - 1) 
LOAN PMT =$ 96.66 - on the 1st loan for $5,000
LOAN PMT =$ 193.33 - on the 2nd loan for $10,000.

8.   Since - 5 is a root.....we canuse synthetic division to find the other two roots

- 5 [ 12     55      -   27       -  10 ]

               -60          25           10        

       ______________________

       12      -5       -2                0

The remaining polynomial is  12x^2  - 5x  - 2

We can factor this as   [ and set to 0 ]

( 4x + 1) ( 3x - 2)   = 0

Setting each factor to 0  and solving for x gives   us the other two roots  as x = -1/4  and x = 2/3

9.       -2, 1  and 5

The polynomial  is  y = ( x + 2) ( x - 1) ( x - 5)

Here's the graph :  https://www.desmos.com/calculator/1fvhn73iyz

OK....not too bad

Rate of change from  x = 1 to x  = 10  =   

[ log 10 - log 1 ]            [ 1 - 0 ]                1

____________   =    _________  =      ____

   10   -   1                       9                       9

Rate of change from 1 to 100

[ log 100 - log 1 ]                2  -  0               2

_______________  =    _________  =   _____

       100  -  1                        99                 99

Since  1/ 9 >  2 /99.....the average rate of change [ slope ] from 1 to 10  is greater than the average rate of change from 1 to 100.....but.....the rate of change between any two points will be positive...so....the slope is always positive

Thank you! 

\(\text{If you have a function }g(x) \text{ which is neither even nor odd (such as }2^x)\\ g_e(x)= \dfrac{g(x)+g(-x)}{2} \text{ is even and }\\ g_o(x) = \dfrac{g(x)-g(-x)}{2} \text{ is odd and }\\ g_e(x) + g_o(x) = g(x) \)

\(\text{Apply this to the two terms of the expression that are neither even nor odd}\)

Midpoint of first segment  =  ( [2 + 0] /2, [4 - 2] / 2 )  = ( 1, 1)

Midpoin of the second segment  =  ( [ 5 + 1] / 2, [1 + 5] / 2 )  =  (3, 3)

Slope  =  [ 3 - 1 ] / [ 3 - 1 ]  =  2 / 2  =   1

Odd, that is what I got aswell but the only answer choices are, .2420, .4840, .0480, .1210
I think I will just go with .2420

C( 19, 3)  * (5/6)^16*(1/6)^3  ≈ .2426

t = -120  =  - 2pi / 3

Sin and cos will be negative (quadrant III)

Cos = - 1/2    (or  0.5)

Sin = - sqrt3/2   (or - 0. 866)

Sum = a₁ / (1-r) 

45 = a₁ /(1- -1/2)

45 (1 1/2) = a₁

_{= 1.5*45 = 67.5}

x/x = 1    I think you meant   x/ (x+4)<----------      and ----------->  (-9/(x+3))

multiply both sides by (x+4)

x = (x+4)(-9/(x+3) )     now multiply both sides by - (1/(x+3)

-x(x+3) = 9x+36

-x^2 -3x = 9x + 36     re-arrange

-x^2 -12x - 36 = 0      or

x^2 + 12x + 36 = 0     factor

(x+6)(x+6) = 0               x = -6   (two roots)

sin(t)=-0.580611184

cos(t)=0.814180971

What is CE?

x + 2y  = 4    (1) 

2x - y  = -7  ⇒   y  = 2x + 7   ( 2)

Sub (2) into (1)  for y

x + 2 [2x + 7]  = 4

x + 4x + 14  = 4

5x + 14  = 4      subtract 14 from both sides

5x = -10            divide both sides by 5         

x = -2

So....using (2).....y  = 2(-2) + 7  =  3

So....the last equation becomes

-2 + 3 + z  = - 4

1 + z  = -4      subtract 1 from both sides

z  = - 5

{x, y, z }  =  { -2, 3, - 5}

Note that triangle BCD is right  ....with legs  BC and DC 

So  DB  =  √[ BC^2 + DC^2 ]  =  √ [ 3 + 9 ]  = √12  = 2√3

And triangle  BEC  is similar to triangle BCD

So....

CE / BC = DC/ BD

CE / √3  = 3 /[ 2√3]

CE   = 3 / 2   =  1.5

EDIT TO CORRECT AN ERROR!!!