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okay so ive been stuck on this question for a while can someone pls help me with this thank you

x  =  number of round tables     and     y  =  number of rectangular tables

Each round table has 6 chairs. Each rectangular table has  10  chairs.

The total number of chairs must be greater than or equal to  185  .

6x + 10y  ≥  185

There must be no more than  15  rectangular tables.

So the number of rectangular tables must be less than or equal to  15 .

y  ≤  15

There must be no fewer than  6  round tables.

So the number of round tables must be greater than or equal to  6 .

x  ≥  6

The number of rectangular tables must be greater than or equal to one-third the number of round tables.

y  ≥  \(\frac13\)x

3y  ≥  x

x  ≤  3y

First box:     x⁴ - 2x²y² + y⁴

Second box:     2x⁴ - 2x²y² + 2y⁴

Third box:     2(x⁴ - x²y² + y⁴)

g(x)   =   2^{x - 3} + 2   =   2^{x - 6 + 3} + 3 - 1   =   (2^{(x + 3) - 6} + 3) - 1   =   f(x + 3) - 1

So

the graph of g(x)  is the graph of  f(x)  shifted  3  units left and  1 unit down.

2.   (tan²x) / (1 + sec x)   =   sec x + 1      Let's test whether this is true when  x = 0 .

(tan²x) / (1 + 1/cosx)   =   1/cosx + 1

(tan²0) / (1 + 1/cos0)   =   1/cos0 + 1

(0²) / (1 + 1/1)   =   1/1 + 1

0  =  2       This is false, so this equation is not true for all values of  x  and so it is not an indentity.

If you want to find the solutions to the equation, we can try to find them.

(tan²x) / (1 + sec x)   =   sec x + 1       Multiply both sides by  (1 + sec x)

tan²x   =   (sec x + 1)(1 + sec x)

tan²x  =  sec x + sec²x + 1 + sec x

tan²x  =  sec²x + 2sec x + 1                Rewrite  tan  and  sec  in terms of  sin  and  cos.

sin²x / cos²x  =  1/cos²x + 2/cos x  +  1        Multiply through by  cos²x

sin²x  =  1 + 2cos x + cos²x                 Subtract  sin²x  from both sides.

0  =  1 + 2cos x + cos²x - sin²x           Substitute  1 - cos²x  in for  sin²x

0  =  1 + 2cos x + cos²x - (1 - cos²x)

0  =  1 + 2cos x + cos²x - 1 + cos²x

0  =  2cos x + 2cos²x         Divide through by  2 .

0  =  cos x + cos²x             Factor  cos x  out of both terms.

0  =  cos x(1 + cos x)         Set each factor equal to zero.

cos x  =  0      or      1 + cos x  =  0

However, neither of these can be true.

If cos x  =  0  then  sec x  is undefined and the original equation is undefined.

If  cos x  =  -1  then  1 + sec x  =  1 - 1  =  0  and the original equation is undefined.

So this equation has no solutions.

1.   sin²(x) - 2  =  -sin(x)

This can't be an identity because it is not always true. For instance...

sin²(0) - 2  =  -sin(0)

0² - 2  =  -0

-2  =  0        This is false, so this equation is not an identity.

However, this equation does have solutions. We can find its solutions.

sin²(x) - 2  =  -sin(x)

                                             Add  sin(x)  to both sides of the equation.

sin²(x) + sin(x) - 2  =  0

                                             Factor the left side like this:  u² + u - 2  =  (u + 2)(u - 1)

(sin(x) + 2)(sin(x) - 1)  =  0

                                                             Set each factor equal to zero.

sin(x) + 2  =  0     or     sin(x) - 1  =  0

sin(x)  =  -2          or     sin(x)  =  1

sin  returns values between  -1  and  1  inclusively, so we can rule out the first option. That leaves...

sin(x)  =  1

x  =  arcsin(1)  =  90°

There will also be a solution every time  360°  is added to  90° , so all the real solutions are

x  =  90° + 360°n      where  n  is an integer

Does this answer your question? I might have misunderstood your question.

thx this is really helpful to my study 

and yes i have tried to solve them i just want to check if i am right because for the first one i got 602.88 but im not sure if thats corect

sorry i know it seems troublesome these are for my study packet 100 pages im trying to study for my math but these questions are just getting to me ive been needing to prepare for the upcoming exam

HI Cphill

I did not really understand exactly second part of question but I have interpreted it as follows:

Define Pattern :

S_{c = 1^{3 +} 3^{3 +} 5³..........+ (2n-1)³  =  n²(2n²-1)  (as per formula for sum of cubes for odd integers)}

Now Sum of cubes for n+(n+1)

S_{2n+1 = n²(2n²-1) + [ 2(n+1)-1]³}

= 2n⁴- n² + [2n+1]³

= 2n⁴- n²  + 8n³+ 12n² + 6n + 1

= (2n⁴ + 4n³ + 2n²) + (4n³ + 8n² + 4n) + (n² + 2n + 1)

= 2n²(n+1)^{2 +} 4n(n+1)² + (n+1)²

⁼ (n+1)²(2n² + 4n + 1)

i understand what i did wrong thanks for the help

Depends on what the value of delta  is .......

You'll just have to use a calculator   to find   cos (-3/2 degrees) = .999657

or if you need   cos ( -3/2 radians) = .070737

You'll need to give alittle more information so we can understand what the issue you are having is...OK ? 

Seriously Sam....you have posted 14 questions...have you TRIED to answer ANY of them?  Or are you just wanting answers to your homework?   Why don't you TRY to answer some and see what you get?  THEN ask a question !

This one:

tan x =  sin x /cos x       cos and sin   at  pi + x  are just the negative of  cos (x) and sin (x) 

tan (pi +x)  =  sin (pi +x) / cos(pi + x) =   -sin(x) /  -cos(x)  = sin(x)/ cos(x)  = 2/5

I'll answer more of your questions after I see your attempts at answers.....Have a good night !       

Definition:  (I got this from Alan the other day....I am learning....are YOU?)

      Angle in radians = arc / radius

15pi/60 = pi/4     AND this is the THIRD time this question has been posted in 3 days!!!!

Second One = FIrst One

=  sqrt ( (24*18*15)*(x*y*y)  )

= sqrt ( ( 2^2 *6  *  3^2 *2  * 15 )  *  xy^2  )

= 2*3*y  sqrt(6 *2*15 x)

= 6y  sqrt( 180 x)

= 6y sqrt ( 36 * 5  x)

= 6y sqrt (6^2 * 5x)

= 36 y sqrt (5x)  

Are you using Khan Academy? You know if you got the question wrong they would give you an answer and explanation.

Factor the following:

-4 y^2 + 3 x y - 12 y + 9 x

Factor terms by grouping. -4 y^2 + 3 x y - 12 y + 9 x = (3 x y - 4 y^2) + (9 x - 12 y) = y (3 x - 4 y) + 3 (3 x - 4 y):

y (3 x - 4 y) + 3 (3 x - 4 y)

Factor 3 x - 4 y from y (3 x - 4 y) + 3 (3 x - 4 y):

=(3x - 4y) (y + 3)

It  is "High Finance", but I will take a stab at it!.

Promissory notes , in general, do not pay regular interest payments, but rather promise to pay the par value of the notes at maturity. The interest rates of the two notes quoted here are for the benefit of the sub-contractor who was owed the PV of these two notes by the ABC company. In other words, instead of paying the sub-contractor the money owned to them, the sub-contractor agreed to receive these two promissory note in lieu of cash payments.

So, to find out how much money was owed to the sub-contractor, will simply find the PV of these two notes at their given interest rates:

PV of $1,000,000 at 7% for 8 years =$582,009.10

PV of $2,000,000 at 5.5% for 5 years =$1,530,268.71

Total PV =$582,009.10 + $1,530,268.71 =$2,112,277.81. This is what is owed to the sub-contractor.

Now, ABC company wishes to replace or exchange these two notes for one note for $3,000,000 par value at 7.5% to mature in 4 years. Ordinarily, the ABC company would simply cancel the two promissory notes in exchange for the new note, with the consent of the sub-contractor. If it is simply a matter of issuing a new promissory note for $3,000,000 for 4 years, it is immaterial whether the note has an attached interest rate of 7.5% or not, from the viewpoint of the sub-contractor, since they will receive $3,000,000 in four years' time. If they wish to calculate the rate of return on their new note, then that works out to 9.17%(since they know the PV and the FV and the term.)

The ABC company theoretically isn't making or losing any money, since they owe the same amount of money, but rather accelerate the amortization of their notes by writing them off a bit sooner. So, with rate of 7.5% over 4 years, the PV of the $3,000,000 note would be booked as $2,246,401.59 for a difference of $2,246,401.59 - $2,112,277.81 =$134,123.78.

Note: All these calculations were done using this online financial calculator: https://arachnoid.com/finance/

Hmmmmm.....not quite.... 

...you got the correct answer by the wrong method

remember

   -1 <= sin <= 1

csc a = 2        1/sin  = 2         sin = 1/2

Since you have the definition of 'y' in the first equation  2+1/4x   let's substitute THAT  for 'y' in to the second equation

3x - 5 (2+1/4x)  = -3     then solve for x

3x - 10 - 5/4  x  = -3

3x-5/4 x = 7

7/4 x = 7

x = 4              Now sustitute this value of 'x' into one of the two equations to find 'y'

3(4) - 5y = -3

12 -5y + -3

-5y = -15

y = 3

The two tan ones are not correct, the others are correct.

f/r^2 +a = n+k            SUbtract a from both sides

f/^r^2  =  n+k - a         Multiply both sides by r^2

f  = (n+k-a) r^2            divide both sides by (n+k-a)

f/(n+k-a)  = r^2             take sqrt of both sides

+- ( f/(n+k-a) ) = r