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Where is angle C? We need Angle C. Once you get angle C, make an equation so \((45-x)+(9x-35)+\angle C=180\), then solve for x and get each unknown angle.

You are very welcome!

:P

Thanks CPhill!

PM:  It may be a form of flattery but it is not very polite to change your answer to match mine.   

If it is  "simple interest" as he/she says in the question, then you don't use exponentiation, which automatically means "compound interest". In this case, you would simply: multiply the principal x interest rate x number of periods, or years in this case. So:

$500 x 0.042 x 4 =$84.00 simple interest earned over 4 years.

$500 + $84.00     =$584.00 principal + simple interest in year 4.

YES, YOU ARE RIGHT !!. 

I've changed my mind a bit.

I think the very first answer might be correct.

There are 7 ways to choose the colour in the middle

Then you can put one colour any other place and there are 5! ways to place the rest.

This allows for rotation but it does not allow for reflectionl

There are 6 axes of symmetry for divide by 6 and you might have athe correct answer.

\((7*5!)/6\)

\(\frac{7*5!}{6} = \boxed{140}\)

Thanks for reminding me of that \boxed command PM 

5. Suppose the polynomial f(x) is of degree 3 and satisfies f(3)=2, f(4)=4, f(5)=-3, and f(6)=8. Determine the value of f(0).

I don't have time to go through this one fully....but...we have this system

27a + 9b + 3c + d =  2

64a + 16b + 4c + d = 4

125a + 25b + 5c + d = -3

216a + 36b + 6c + d = 8

a = 9/2

b = -117/ 2

c = 245

d = - 328

The polynomial is

(9/2)x^3  - (117/2)x^2  + 245 - 328

So

f(0)  =  d    =   -328

NOTE    :    #1   is solved similarly.....6 equations....6 unknowns.....a computer algebra system may be the fastest way to go......that's a lot of work to solve such a large system by hand....!!!

See if you can do it based on this one

There are seven colours MP

If you are not going to consider any over counting then there are 7! ways to put thise in place.

Where do you get 6!  from? 

If you were going to put 7 coloured beeds in a circle (rotations classed as the same) then the answer would be 6!

But this is not a circle AND you said that you were counting each rotation as different for this part of the question. 

4. Find an ordered pair of constants (a,b) such that the polynomial f(x)=x^3+ax^2+(b+2)x+1 is divisible by x^2-1.

If the polynomial is divisible by  x^2 - 1

Then it is divisible by    (x -1)  (x + 1)

This means that     x = 1     and   x  = - 1    are roots

Therefore

(1)^3 + a(1)^2 + (b+2)(1) + 1 = 0    →   1 + a + b+2 + 1 = 0 →  a + b = -4     (1)

And

(-1)^3 + a(-1)^2 + (b + 2)(-1) + 1 = 0 → -1 + a - (b + 2)  + 1 = 0  → a - b = 2   (2)

Add (1)  and (2)  and we get that

2a = -2

a = -1

And

a - b = 2

-1 - b = 2

b = -3

The polynomial is

x^3 - x^2 - x + 1

3. Find a monic quadratic polynomial f(x) such that the remainder when f(x) is divided by x-1 is 2 and the remainder when f(x) is divided by x-3 is 4.

Let the polynomial be    x^2 + bx + c

Use synthetic divsion

1     [    1      b               c    ]                      3  [  1         b          c    ]

                    1             b + 1                                        3       3b + 9              

         __________________                           ___________________

           1     b + 1    [1 + b + c]                          1       b+ 3   [ 9 + 3b + c]

We have this system

1 + b + c  =   2           ⇒  b + c   = 1

9  + 3b + c  =  4     ⇒   3b + c  =  -5

Subtract the 1st equation from the second and we get that

2b  =  -6

b = -3

So

b + c  = 1

-3 + c = 1

c = 4

So....the polynomial is

x^2 - 3x + 4

OK....glad to help   !!!!

I only have time for a couple.....

2. 6 is the remainder when g(x) = x^9 - k x^5 + 4 is divided by x - 1. Find k

Using synthetic division and accounting for missing powers, we have....

1     [    1     0     0     0   - k       0        0       0      0      4  ]

                   1     1     1    1      1- k     1-k   1-k   1-k   1-k

           ________________________________________

             1     1    1     1   1-k   1-k      1-k    1-k  1-k   5-k

5 - k   is the remainder

So

5 - k  =  6    

So

k = -1

Yes thank you so much!!! 

Good deal....!!!

Did you understand my answer to your other problem????

Okay Thank you so much!!! This makes so much more sense now!!!!

OK....I know this is confusing    [ don't worry..I'm a "night owl"...LOL!!!]

Note

Tan x  =  14.6/ 7.2       

This tells us the ratio of the opposite /adjacent  sides to the angle , x

The inverse actually tells us what x is   [ the angle measure of x]

So

atan (14.6/ 7.2]  =  x  = 63.7

Remember....the trig function tells us a ratio  ......the inverse returns the angle value of that ratio

Not that difficult.....

We know an opposite side to the angle [ the height of the pole]

And we know the adjacent side to the angle [ the shadow length ]

"x" is the elevation angle [ what we are looking for ]

So....the trig function that relates  opp / adj    is the tangent

So....we have

tan x  =   38 /53

Now....to find the angle use the tangent inverse 

The other problem I anwered went through this step-by-step.....let me know if you didn't understand it......

So

arctan (38/53)  ≈  35.64°  =  36°   =   x

How did you get the x alone? by subtracting, adding, multipling, or dividing tan on both sides? And by doing that it gets you the inverse of Tan which is Tan^-1?

* Thank you for helping me, Ik it's super late.

You're on the right track....just a calculator problem....

On the home page calculator....make sure you're in DEGREE Mode....[at bottom left ]

Key in 

14.6 / 7.2  

Then....hit the  "2nd button"

You will see the  "atan" button....hit that

Then hit "="

You should see   "63.749....."

This where   "63.7"   comes from.....

Hope that helps.....incidentally..."atan"  is the same as "tan^-1 ".....it means "tangent inverse"

ywsm (you welcome so much)

\(6x^2 = 2x-1\).

Subtract both sides by 2x - 1 to get 6x^2 - 2x + 1 = 0. Using the quadratic formula

(\(x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\)) with a = 6, b = -2, and c = 1, we have \(x = \dfrac{-(-2)\pm \sqrt{(-2)^2-4(6)(1)}}{2(6)}\), which simplifies into \(x = \dfrac{2\pm \sqrt{4 - 24}}{12} \Rightarrow x = \dfrac{2\pm \sqrt{-20}}{12}\). Our answer is \(x = \dfrac{2\pm \sqrt{-20}}{12}\), which is the same as \(\boxed{ x = \dfrac{1\pm i\sqrt{5}}{6}}\) . Note that \(i = \sqrt{-1}\).

Hope this helps, 

- PM

Understand that absolute value makes all numbers inside it non-negative (0 or greater). If the x in f(x) is non-negative, then y must be non-negative because f(x) = y.

The graph will look the same, but whenever y < 0, the y value changes to \(|y|\) .

For example, if the coordinate was (-1, -1), the y-coordinate turns into 1 by absolute value. The point on the graph changes to (-1, 1). The graph will not have any coordinates below the x-axis.

Hope this helps,

- PM

The formula for loans is \(x\cdot (1+r)^y\).

x represents the initial loan amount. In this example, x = $500.

r represents the interest rate, whether it is simple or compound (for compound, there is a slightly different formula). in this example, r = 0.042. 

y represents the number of years the loan is not paid back yet. In this example, y is 1, 2, or 3.

- PM

The goal of counting arrangements: Count everything ONCE and ONLY ONCE. 

Understand that once we choose the outer six, the center dot is automatic. 

There are 6! ways to put the colors in any arrangement without considering overcounting. 6! = 720. 

We must consider overcounting. Since there are 6 axes of symmetry, we divide 720 by 6 to get 120. The rotations of the arrangements also matter (Eg: [1,2,3,4,5,6, 7 in the center] is the same as [4,5,6,1,2,3, 7 in the center]), so we divide 120 by 2 to get 60. There is your answer, \(\boxed{60}\).

Hope this helps, (do not get confused by the answers above)

- PM

mean is the same thing as average. Add up all terms, and divide by the number of terms.

if there 3 terms of x, y, and z, the mean or average is \(\dfrac{x+y+z}{3}\).