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Here's a more detailed explanation :

https://web2.0calc.com/questions/i-need-helping-graphing

Answered here :

https://web2.0calc.com/questions/graph-function-over-interval

See here :  https://www.desmos.com/calculator/urgcqnke2a

{ Both lines should have "arrows" at their ends ]

Don't know a way to solve this algebraically...but...by "intersection of two grids" I think you mean "laatice points" [ points with integer coordinates ]

We can construct a a circle with a radius of sqrt (20)  centered at (1, 1)

I found these two possibilities for B....[ and there may be more ]  ....(-3, 3)  and (5, -1)

See the graph here : https://www.desmos.com/calculator/weqndpih8n

The one on the lower left represents the data given.

The sugar is   6   out of (1+9+6) PARTS   = 6/16 PARTS

     Increasing the sugar to 8 cups   8/6 * 16 = 21.33 cups total   = 21 1/3 cups

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2^3*10^3 = 8000

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2 x 10^20   *  3 x 10^17  = 6 x 10^ 37

10 pounds / 3/5 pound/bowl = 16 2/3 bowls ~~~  16 whole bowls.

OK....we have

[ f(t + h ) - f(t) ] / h  =

[ 4 / (t + h)^2   - 4/t^2 ] / h  =  

[  4t^2  - 4 ( t + h)^2 ] / [  h * t^2 * (t + h)^2 ]  =

[ 4t^2 - 4t^2 - 8ht - 4h^2 ] / [ h  * t^2 * ( t^2 + 2ht + h^2 ] =

[  h ( -8t - 4h) ] / [ h * ( t^4 + 2ht^3  + t^2h^2]  ....   [ divide out h ]

[ - 8t - 4h ] / [ (t^4 - 2ht^3 + t^2h^2 ]          let h →  0    and we have that

[ -8t ] /  [ t^4 ]  =

-8 / t^3     =  -8 t^(-3)

Don't worry....shortly.....you will learn to get to this result much faster  !!!!

(5x5) / 6 + (4x9) / (7x4) (Sqrt8) x 12 x 1,000,000

(25/6) + (36)/(28) x (2.828...) x (12 000 000)

(4.1666666667 + 1.285714...) x (2.828.. x 12 000 000)

(5.452...) x (33941125.4969)

185 059 946.162 is the answer!

You do know there is a calculator on this site...yes?  

Take the percentage of it not raining over all the 3 days and times them together to get the chance it will not rain over all those days.

(0.6 * 0.5 * 0.2) = P of rain over all the days

0.06 or 6 percent that it will rain over all those days.

Taking the chance of it raining Monday is:

 100 - 50 = 50 percent chance over monday.

Taking the percent chance of it raining over monay and tuesday:

0.6 * 0.5 = 30 percent chance it will rain over those two days.

Taking the percent chance of it raining over all the days is:

0.6 * 0.5 * 0.2 = 0.06 so a 6 percent chance over all the three days.

Does this make since? 

We have a right triangle...we know the hypotenuse and want to know the side adjacent to the known angle....so....the cosine function seems appropriate...

cos 42 = b / 63.2

63.2 cos 42  = b ≈ 46.97 units

OK....

Note that   3sqrt (2)  =  = sqrt (9 * 2) =  sqrt (18)

And 2sqrt (5)  = sqrt (4 * 5) =  sqrt (20)

So....we can find the height o the triangle by the Pythagorean Theorem...

h ^2  = [ sqrt (20) ] ^2 - [sqrt (18) ] ^2 

h^2  = 20 - 18

h^2 = 2

h = sqrt (2)

And the base = 3sqrt (2) + sqrt (2)  = 4sqrt (2)

So....the area  =  (1/2)base * height =  (1/2) [ 4 sqrt (2) ] [ sqrt (2) ]  =  (1/2) (4) [sqrt (2) ]^2  =

(1/2) (4) (2)  =

4 cm ^2

If the slope of m = 0....the line is horizontal [ parallel to the x axis ]

If n is perpendicular....it is a vertical line [ parallel to the y axis ].....and the slope of a vertical line is undefined

I'm confused on how you got the derivative.  I used f'(x)= [(x+h) - f(x)] / h   and I just have no clue what to do

Confirmed! 

\(\text{Draw a straight line through the centers of the top two circles ending at the rectangle's edges}\\ \text{It should be clear that the length of this line is }4r = 8mm \\ \text{and that it is the length of the horizontal side of the rectangle}\)

\(\text{Now draw a line from the center of the top left circle to the center of the bottom circle}\\ \text{It's length is }2r=4mm\\ \text{The vertical distance between the centers is then }4 \cos(30^\circ) = 2 \sqrt{3}mm\\ \text{You can show this by noting the triangle connecting the centers is equilateral and using a bit of trig}\)

\(\text{The distance from the top of the rectangle to the top left circle center is }r\\ \text{as is the distance from the bottom to the bottom circle center}\\ \text{Thus the vertical side of the rectangle has length }\\ 2r + 2\sqrt{3} mm =4+2\sqrt{3} mm\)

\(\text{The area is just the product of these two numbers}\\ A = (8mm)(4+2\sqrt{3}mm) = 32+16\sqrt{3} ~mm^2\)

The width is no problem.....it is just 4 times the radius of each circle = 4r = 4(2mm) = 8mm    (1)

Note that joining the 3 radii will form an equilateral triangle  with sides = 2r  = 2 * 2mm = 4mm

The height of this triangle, h, can be found as

tan 60 = h/2

sqrt (3) = h / 2

2sqrt(3) = h

So....the height of the rectangle =  [ 2sqrt(3)  +  2r ]  mm   = [ 2sqt(3) + 2(2)] mm  = [ 2qrt(3) + 4 ] mm  (2)

So....the area of the rectangle =  (1) * (2)  =

[ 8 ] *  [ 2sqrt (3) + 4 ]  mm^2 =

 [ 16sqrt (3) + 32 ]  mm ^2