All Answers 358108 Answers

Avg speed  over the same  distance    =  2 * product of the rates /  ( sum of the rates)  =

2 (30)(60) / ( 30 + 60)

I'll leave the rest for you.......

Area of a circle  = 

3.14  * ( diameter / 2)^2

Sum of first  n positive even integers  =   n  (n + 1) 

So

n ( n + 1)   =  2550

n^2 + n - 2550  =  0      factor as

(n + 51) ( n - 50)  = 0

The second factor set  = 0   is what we need

n  - 50   = 0

n  = 50

See my answer under your original post....

Let the original number   = N

So

(5N)  / (N + 7)  = 8

5N  =  8 (N + 7)

5N  = 8N + 56

-56  =  8N - 5N

-56  =  3N

-56/3  =   N

I will start you off.

\(\dfrac{7x^{2014}}{4x^{2013}}\cdot{\dfrac{2x^{2012}}{2x^{2007}}}\cdot\dfrac{27x^{2006}}{x^{2002}}\cdot\dfrac{6x^{2008}}{8x^{2010}}\)

\(=\dfrac{7x^{2014-2013}}{4^{2013}}\cdot{\dfrac{2x^{2012}}{2x^{2007}}}\cdot\dfrac{27x^{2006}}{x^{2002}}\cdot\dfrac{6x^{2008}}{8x^{2010}}.\\ ~\\=\dfrac{7x^{1}}{2^{2*2013}}\cdot{\dfrac{2x^{2012}}{2x^{2007}}}\cdot\dfrac{27x^{2006}}{x^{2002}}\cdot\dfrac{6x^{2008}}{8x^{2010}}\)

Now it is your turn.

Call Ian's rate  =  R

And call the time he drove in hours  ,T

So

Han drove   at  (R + 5)  for a time of (T + 1)

And

Jan drove  at (R + 10)  for a time of (T + 2)

We have  this  equation

(R + 5) (T + 1)  -  RT  =  70

RT + 5T + R  + 5  - RT  =70

5T + R  =  65

R =  65 - 5T

So  the distance that  Ian drove =  RT  =   (65 - 5T)(T)   =  65T - 5T^2      (1)

And the distance  that Jan drove  =  (R + 10) (T + 2)  =   (65 - 5T + 10) (T + 2)  =  (75 - 5T)(T + 2)  = 

75T - 5T^2 + 150 - 10T  =   65T - 5T^2  + 150    (2)

Subtract  (1)  from (2)

65T - 5T^2 + 150  = 65T - 5T^2  =   150

So  Jan drove 150 miles farther than Ian

8^(x^2)*4^x = 2

We can write

(2^3)^(x^2)*(2^2)^x  = 2^1

[2^(3x^2) ] * [ 2^(2x) ]  =  2^1

2^ [ (3x^2 + 2x) ]  = 2^1             we have the bases equal....solve for the exponents

3x^2 + 2x  =  1      complete the square

3(x^2 + (2/3)x + 1/9)^2  = 1  + 1/3

3 ( x + 1/3)^2  = 4/3                      multiply both sides by 1/3

(x + 1/3)^2  =  4/9              take both roots

x + 1/3 =   2/3                or        x + 1/3  =  -2/3

x   = 2/3 - 1/3                            x = -2/3 - 1/3

x = 1/3                                      x =  -1

csc B  =   hypotenuse /  opposite   =   65 / 33

What is cos Z?

Hello Guest!

\(z:y=2:1\\ x:y:z=\sqrt{5}:1:2\\ \color{blue}cos\ z=\frac{1}{\sqrt{5}}=\frac{1}{5}\cdot \sqrt{5}\)

  !

(a) Let us say potatoes are equal to x

     carrots are equal to y.

     Then the question we know

                x + y = 250 - (i)

                x + 20 = y - (ii)

(ii) equation 

         x = y - 20 put this in eq (i)

          y - 20 + y = 250

               2y = 250 + 20

               2y = 270

                 y = 135

put this in eq (i)

x + 135 = 250

x = 250 - 135

x = 115

x = 115, y = 135

So he ordered 135 carrots

(b) Let us say the chef used 3 carrots

    We can get

= 115 - 3y = 1/2 (135 - y)

= 115 - 3y = 1/2 * 135 - 1/2y

= 115 - 3y = 67.5 - 1/2y

= 115 - 67.5 = -1/2y + 3y/1

= 47.5 = (-1 + 6)/2 y

= 47.5 = 8/2y

= 47.5 = 2.5y

= 47.5/2.5 = y

= 19 = y

y = 19

so 

115 - 3 * 19 + 115 - 57 = 58

So 58 potatoes were left

When multiplying numbers with the same base, add their exponents.

To add  1/3  +  1/9  +  1/27  +  ...

note that it is an infinite geometric sequence whose sum is:  Sum  =  a / (1 - r)

where   a  =  1/3     < the initial term >

and       r  =   1/3    < the common ration >             so:            Sum  =  (1/3) / [ 1 - (1/3) ]

                                                                                                Sum  =  (1/3) / (2/3)

                                                                                                Sum  =  1/2

So:  the product  =  2^1/2  =  sqrt(2)

Add all the equations, then divide by 10.

The solution is [0,1/2).

Do not post A O P S homework.

We are in Quad 4

r =  sqrt [ 9^2 + 4^2 ]   = sqrt  97

a)   sin =   -9/sqrt 97

      cos =  4/sqrt 97

      cot = -4/9

       sec =  sqrt (97)  / 4

       csc =  -sqrt (97) /4

b)   arctan (-9/4)   =  -66.04°

So....the angle is  360  -  66.04  =  293.96°

a)    r = 10   x = -6  y  =  -8

sin A  =  -8/10  =   -4/5  

cos A =  -6/10 =  -3/5

tan A  =  -8/-6 =  4/3

cot A =  3/4

sec A  =  5/-3

csc A =  -5/4

b)   arcsin (-4/5)  = -53.1°

However, this  angle lies in Quad 3   because both x and  y are  negative

So....the angle is  180 + 53.1 =  233.1°

y^2 - 4y + 4  factors as    (y -2)^2

So

(y -2)^2 = 0

So  when y = 2, the denominator   = 0  and  the expression is undefined