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The first one is close

We have the form     y = Asin (Bx) + C

A = the amplitude  = 3

We can find B  thusly:

B = 2pi / period =  2pi / 4  =   pi/2

C = the shift up/down ..... the normal midline for the sine is y = 0

This is shifted down 1 unit...so  C = -1

So....our function is

y =  3sin (pi/2 x) - 1

Here is the graph : https://www.desmos.com/calculator/5s6jufmy5k

√ [8x^3y^8 ]  =  

√[ 2^3 * x^3 * y^8] = 

(2^3)^(1/2) * (x^3)^(1/2) * (y^8)^(1/2) =

(2)^(3/2) * (x)^(3/2) * y^4

As before

3/2 = 1 + 1/2         1 power of 2 comes out 1 stays in

3/2   = 1 + 1/2        1 power of x comes out, 1 stays in

4 = 4                     4 powers of y come out......0 remain in

So we have

2xy^4√ [ 2x ]

sumfor(n, 1, 2024, sqrt(45 + sqrt(n)) / sqrt(45 - sqrt(n))=7098.9283254388............  

the answer is not 1/5 btw

I find it easiest to put these in expnential form  and then simplify

√  27x^11y^6  =     

(3^3)^(1/2) * (x^11)^(1/2) * (y^6)^(1/2)  =   

(3)^(3/2) * (x)^11/2 * (y)^3

Now.....we can  find out how many powers of  each thing come out of the radical and how many powers stay in by this procedure :

Divide 3/2  and express as a mixed number =  1 + 1/2

The whole number tells how many powers of 3  come out of the radical =  3^1

The numerator tells how many powers of 3 stay in the radical = 3^1

Similarly

11/2 =   5 + 1/2

5 powers of x come out......1 power stays in

And

3  =  3

We have no mixed number.....so 3 powers of  y come out   and there remain 0 that stay in

So....we have

3^1* x^5 * y^3 √  [ 3^1 * x^1 ]  =

3x^5y^3 √ [ 3x ]

sumfor(n, 1, 9800, (1/(sqrt(n+sqrt(n^2 - 1)))) = 139.2964645562 ...........

Each leg will  =      hypotenuse length / √ 2

Here's the graph :  https://www.desmos.com/calculator/gvm2oy7mhe

a) Domain = [ 0, infinity )

b) Range =  [ 0, infinity )

c) We cannot take the square root of a negative  (it is not a real number)

d) We can only get 0 or a positive number out of a positive root

e)  Avg rate of change   =  [ √ 16 - √ 1 ] / [ 16 - 1 ]  = [ 4  - 1 ] / 15   =  3/15    = 1/5 

f)   the function is  g(x) =    √ ( x + a)  + b

We can answer this with Calculus....but a graph seems better : https://www.desmos.com/calculator/pacud9kcer

We have two x values that minimize  (x + 1) (x + 2) (x + 3) ( x + 4)

They are 

x ≈ -3.618    and   x ≈ -1.382

And the minimum value is y = -1

So....these  x values also minimize   (x+ 1) (x + 2) ( x + 3) (x + 4) + 2019   =  -1 + 2019 = 2018 

square both sides to get:

x^2-4x-5 = 4x^2 -40x + 100      re arrange terms to simplify

        0 =  3x^2 - 36x +105    divide by 3 to simplify

        0 =  x^2 - 12x +35        factor to solve

      0 = (x-7)(x-5)               x thus equals   7   and  5    

y = 2 + √ [ y^2 - 12 ]         subtract 2 from both sides

y - 2 =  √ [y^2 - 12 ]         square both sides

( y - 2)^2 = y^2 - 12

y^2 - 4y + 4  = y^2 - 12       subtract y^2 from both sides

-4y + 4 = -12        subtract 4 from both sides

-4y = -16       divide both sides by -4

y = 4

Check that this solution works in the original equation, GM....

 x + y = 12

x - y =  18       add these

2x = 30       divide both sides by 2

x = 15

And using    x + y = 12    we have

15 + y = 12       subtract  15 from both sides

y = -3

So  (x,y)  = (15, -3)    ⇒  the smaller number is -3

Something is deleted.

The shortest distance on the surface will be along the shortest dimension and then the diagonal of the rectangle

replacing 10 12 20 with   12  16  36 respectively   travel from A along 12cm edge then diagonal of  16 x 36 =4 sqrt97

4 sqrt 97  +  12    cm

If we join the centers of the top three circles....we will have an equilateral triangle with a side of 8

The height of this triangle  4√ 3  units

And if we join the bottom three circles...we have another equilateral triangle with the same height

So....by symmetry......the total height  =  2r + 2 (4√ 3) =    2(4) + 8√ 3  =

8 + 8√ 3 =

8 ( 1 + √ 3) units

Why does it show, "Waiting for moderation"? Hopeully, one of the mods could approve it.

Here it is:

The ENTIRE line is a common point....it is one line on top of the other, so there is no 'point' of intersection....all of the points on the lines are called 'collinear' rather than intersection.

Yikes....that is a bit garbled...I'm going to say the area is 15 sq ft   and the width is 3

Area = w x h      find h to see if it will fit

15 = 3 x h

h = 5      it will just fit !  

Recognize that the sqrt function is the same as raising something to the  1/2 power

1st one     sqrt( 50r^2s^4) = (50r^ s^4)^1/2   =  r^(2*1/2) s^(4*1/2)  50^(1/2)   = r s^2 sqrt(50)  = r s^2 sqrt(25*2) = 5 r s^2 sqrt (2)

   DO you see?    Try the next one...

       x^(3*1/2)   realize = x^3/2 =   x  sqrt(x)

First one

Consider    2^x

When x =  0      y = 2^0  = 1

So...the point (0, 1)  is on the graph of g(x)

The only graph on which this happens for (x) is the first one

So the solutions are x = 0   and x = 2

For the second one we have only one intersection point  ... x = 1

Sorry for the late reply. Thank you so much for your help. I don't use desmos very often so I didn't even think about using it!

2x + 6y= 18

-3x - 9y= -27

To make this easier  divide the first equation through by 2    and the second through by 3 and we have that

x + 3y   =  9

-x - 3y  = - 9          add these

0 = 0

When we have this situation.....there are infinite solutions  for (x, y)

For instance   (0, 3)  solves the equations

or

(9, 0)  also   works

This is actually the same line....mutiply the second equation by -1  and we get

x + 3y = 9       the same as the first equation  !!!

Thanks! it does. would you mind checking my work on my most recent question on this