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let the first integer be n

the second is (n+1), the third (n+2)

4n = (n+2) + 13

4n = n + 15

3n = 15

n = 5

4 x 5 = 20 = 7 + 13

the 3 integers are 5, 6, 7

D. 36

Let the portion Jenny can eat in one hour  = 1/x

Let the portion  that  Penny can eat in  one hour = 1/y

Let the portion that Lenny can eat in one hour = 1/z

And we have this system

2/x + 2 / y  = 1      (1)

(5/3)/y + (5/3)/z =  1      (2)      

1/x + 1/y + 1/z  = 1   (3)

Multilpy (2)  by  (3/5)  and we have

1/y + 1/z  = 3/5     (4)

Put (4) into (3) and we have

1/x + 3/5  = 1     subtract   3/5 from  both sides

1/x = 2/5  ⇒   x  = 5/2  = 2.5 hrs = time for Jenny to eat the whole gallon by herself

Multiply the first equation  by  -2  and we have

-2x  - 2y  - 6z  = -20     

-4x  + 2y + 5z  = 7          add these  and we get

-6x   - z    =  -13     (1)

kx  +   z   =      3

Multiply (1)  by - 1

6x + z  =  13

kx + z  = 3

This  system will have no solution  when k  =  6

(A)  is true

(B)  is true

(C) is true

Thanks

2. The cos is  > 0   from  [0, to pi/2)   and from (3pi/2,  2pi ]

    The csc  is < 0  where the sine is < 0....from  (pi, 2pi )

So the cos > 0  and csc < 0   from  (3pi/ 2, 2pi )

3.  cos (theta)  = - sqrt (3) / 2       at    -5pi/ 6   and at  7pi/ 6

1)  Distance <  Time * Rate      ...so....

D < 2 (17.6)  ⇒   D < 35.2

2)  Multiply  6.5 * 17.6  =  114.4 miles..since this is greater than 112 miles...you will be able to finish...

did you try dividing? no.. as you clearly tried nothing 

\(5x^2 + 10x - 15 = (x+5) 5x -(15x+15) \\ -(15x+15) = -15(x+5) +60 \\ 5x^2 + 10x - 15 = (15x-15)(x+5) + 60 \\ \text{The remainder is 60}\)

Sorry, A and C is the wrong answer.

Sorry, I already tried 3; it is the wrong answer.

Correct!!!

[90 + 95 ] / 2  = 92.5

Good job!!!

Yep...!!!

Correct...!!!

OK...no prob  !!!

Using  I  = Prt

Where P  = 2900  r = 3.35%  = .0335   and  t  = 1

Interest charge at the end of the first month = 2900 * (.0335)   = $97.15

hey thanks man for the fast reply you're a big help!

27 * 9^(3x)  =

3^3  * (3^2)^(3x)  =

3^3 * 3^(6x)  =

3^ (3 + 6x)

So...k  = 3 + 6x