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For  a function, we cannot  two  diferent  y  values associated with the same x  value

So..

A  is  OK

B is not

C is OK

So  A  and C

Say the number is ABCDE

"A" would have 9 solutions 1-9

"B" would also have 9 solutions 1-9 - one number +zero

"C" would have 8 solutions

"D" would have 7 solutions

"E" would have 6 soultions

9*9*8*7*6 is 

27216

:)

Find the greatest integer value of b for which the expression (9x^3+4x^2+11x+7)/(x^2+bx+8) has a domain of all real numbers.

Domain is all real numbers iff x^2+bx+8 ≠ 0.

Let x^2 + bx + 8 = 0 and then we will find a discriminant that will result in no real solutions if x^2+bx+8 = 0 because the denominator ≠ 0.

A equation is unreal iff b^2 - 4ac < 0

So b^2 - 32 < 0

b^2 < 32

floor(sqrt(32)) = 5

5 is the answer I think

Np! and dnt mention it

Thank you!

The consecutive angles of a particular trapezoid form an arithmetic sequence. If the largest angle measures 120 degrees, what is the measure of the smallest angle?

Let a be the smallest angle

Let n be the difference between two consecutive angles

a+(a+n)+(a+2n)+(a+3n)=360 (1)

a+3n=120 (2)

Substituting, we have: a+(a+n)+(a+2n)+120=360

a+a+n+a+2n=240

(a+a+a)+(n+2n)=240

3a+3n=240

a+n=80 (3)

(3) * 3 = 3a + 3n = 240 (4)

(4) - (2) => 2a = 120

a = 60 degrees

60 degrees

From what I know it is always a constant but if you do have a remainder u have to make that a remainder a constant so in terms, u can very well get a remainder but you have to make it into a constant in order for it to work

f(x) = 2/3 x + 3

x=12

f(x)=?

f(x) = 2/3 * 12 + 3

f(x) = 4 * 2 + 3

f(x) = 8 + 3

f(x) = 11

Fixed fee is $5.75, since it is an initial fee. It would be torture if fees weren't fixed.

Look whoever you are number one I needed help because that question was for a high point assignment and #2 don't assume things I'm not asking to be of main priority just asking for help so mind your business and stop giving people lip its rude you don't need to start things with people because of a question title that's unintelligent of you

If 2x - 9y = 14 and 6x=42+y, what is the value of the product xy?

2x - 9y = 14 (1)

6x = 42 + y (2)

(1) × 3 = 6x - 27y = 42 (3)

(2) & (3) = 42 + y - 27y = 42

y = 27y

Only holds if y = 0 assuming y is a real number.

If x is also a real number, x * y will always be 0 because y = 0.

xy=0

Since a trapeziod is a quadrilateral, the sum of the angles is 360 degrees. Let x represent the difference between each term in the arithmetic sequence. This means that 120+(120-x)+(120-2x)+(120-3x)=360. Solving for x, you have that the common difference of each term in the arithmetic sequence is 20. The smallest angle is (120-3x), so substituting 20 for x, you get 120-3*20=60. 

The measure of the smallest angle is 60 degrees.

Note  that  the second equation can be arranged as

y =  6x  -  42

Sub this into  the first equation for  y

2x - 9 ( 6x - 42)  =  14       simplify

2x  - 54x  + 378  =  14

-52x  =  14  - 378

-52x  = -364

x  = -364 / -52    =   7

And   y   =  6(7) - 42  =  0

So....xy  =  7*0  =  0

When it hits the ground, y =   0...so....

We  just  need  to solve this

-4.9t^2  + 42t  +  18.9   = 0 

Using  the quad fromula  we get

[ -42 ± sqrt  (42^2  - 4 ( 18.9) ( -4.9) )]  / ( 2 * - 4.9)   

Taking the positive  value.....t  =  9   

( It  hits  the ground at 9 sec )

q   must  be   negative  ( if it  were positive, we  would  have factors of ( x + m) ( x + n)  =  0, and the roots  m, n  would both be negative)

So......any  negative for q  would  work  because  the discriminant

3^2  - 4(q)(1)    will  be  positive

So

q  <  0

Thank you for the first answer but can someone please answer the 2nd question?

Yea, you were correct, Cubed root 3 was odd, not neither. The answer was 1,3,4

ok thanks, i'll check

1.      (x + 2)  / ( (2x + 2)   =   (4x + 3)  / ( 7x + 3)      cross-multiply

(x + 2) ( 7x + 3)  =  ( 2x + 2) ( 4x + 3)     simplify

7x^2 + 17x  + 6   =  8x^2  +  14x   + 6            

7x^2  + 17x  = 8x^2  +  14x      rearrange   as

x^2  -  3x  =  0

x ( x -  3)  =  0

x = 0  is one solution.....so....the product   of all real solutions  must  be    0

I think  \(y=\sqrt[3]{x}\)  is not neither.

\(f(x)=\sqrt[3]{x}\\~\\ f(-x)\ =\ \sqrt[3]{-x}\ =\ \sqrt[3]{-1\cdot x}\ =\ \sqrt[3]{-1}\cdot\sqrt[3]{x}\ =\ -1\cdot\sqrt[3]{x}\ =\ -\sqrt[3]{x}\ =\ -f(x)\)

(It is a little confusing because when I check that in WolframAlpha, it only agrees when it uses the "real-valued root")

Otherwise I agree with your answers

See  here :   https://web2.0calc.com/questions/i-m-a-little-confused

No prob  !!!

The denomiantor cannot  =  0 

So...we can guarantee  no  reals  will make the  denominator =  0   if we solve this :

b^2  - 4 (8)(1)  <  0

b^2  - 32  <  0

b^2  <  32        take  the positive root

b  < sqrt 32

b < 5.6

So  5  is  the  greatest  integer

Thank you!

The common difference is    -1/3 - 1/4   =  -7/12

We need to evaluate this

-1/3  + (7/12) ( 10 - 2)

-1/3  + (7/12) ( 8)

-1/3  +  56/12

-1/3  +  14/3  =

13 / 3   =  the  second term