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\(If\\ y=log_{10}100 \qquad then\\ 10^y=10^{log_{10}100}\\ i.e.\\ 10^y=100\\ y=2\)

Allowing for repeats, we have

80, 81, 82, 83 , 84, 85, 86, 86, 87, 87, 88, 88, 89 , 89 , 90, 90 , 91, 91,

92, 92, 93, 93, 94, 94, 95, 96, 97, 98, 99, 100

We' ll  give the answer in mm^3

Note.....1 m   =   1000mm      and 10m  =  10000 mm

The volume  of the tank (without the paint)  =   

(4/3) pi (1000mm) ^3  +  pi (1000mm)^2 (10000mm)         (1)

The volume of the tank (including the paint)  =

(4/3)pi (1001mm)^3  + pi (1001mm)^2 (10000m)       (2)

So.....the volume of paint required   =  (2)  - (1)  =

(4/3)pi [ 1001^3 - 1000)^3]  +  (10000)pi [1001^2 - 1000^2] =

(12012004 π)/3  + 20010000 π    ≈  75,442,210  mm^3

Wavelength =Speed of light / Frequency

W =3E8 / 5E14

W =0.0000006 =6E-7 meters - the wavelength of the radiation in the optical environment,

The "complement" of a set just means all the elements of the universal set, U, that are not found in that set......so.....

The complement of A, denoted as A' =  { 2,3,4,6,7, 8, 9 ,10}

The complement of B  = B' = {1,2,6,7,9,10,11,12}

The complement of C  = C'  { 1,3,4,5,6 }

The  union of the complement of A, the complement of B, and the complement of C  means a listing of all the elements  in either A' ,B' or C'  or all three.

So  using the symbol U to denote union, we have

{ A' U B ' U C ' }   =   { 1,2,3,4,5,6,7,8,9,10,11,12 }

So....finally.......we want to find the complement of this last set......this means finding all of the elements in  the universal set  that are not contained in this set......notice that there are none.....so....the result of  this is called the empty set, denoted as  {    }

And that's the final answer

2e^-0.07t=8   divide both sides by 2

e^ (-.0.07t)  = 4     take the Ln of both sides

Ln e^(-0.07 t)  = Ln 4      and by a log property we can write

-0.07 t    = Ln 4             divide both sides by -0.07

t =  Ln 4 / -0.07 ≈  -19.8

I assume that you want to solve for x

So we have

tan (70)  =  x / 7      multiply oth sides by 7

7 *  tan (70)  =  x   ≈ 19.23

Excellent explanation, Guest.....why don't you register and become a member???

We need more people like you on here!!!!

You don't have to graph this to figure out the answer. Just know the relationships (although I prefer graphing than remembering the relationships.

A reflection over the x-axis has the following effect:

\((x,y)--->(x,-y)\)

A reflection over the y-axis has the following effect:

\((x,y)--->(-x,y)\)

Let's try this with point C at (2,2):

First, you reflect over the x-axis. This arrow notation says that the x-coordinate does not change but the y-coordinate changes to -1*y-coordinate:

\((2,2)--->(2,-2)\),so \(C'(2,-2)\)

Take C' and apply the effect of reflecting over the y-axis:

\((2,-2)--->(-2,-2)\),so \(C''(-2,-2)\)

Hopefully, this makes sense. This is simply another method of solving. If you want to stick to graphing, that is OK.

21!!!

Okay, thanks!

\(\lim_{x\rightarrow 2} [(χ^2-5χ+6)/(χ-2)] \)Can you find it?(it's easy i trust you) 

The formula for finding the area of a parallelogram is:

Let A= area

Let b= base

Let h= height 

\(A=bh\)

If f(x) = Log(x),

then, f ^-1(x) = AntiLog (x).

Also, AntiLog (x) = 10x

\({8}^{2}=64 \) 

So 8 is the square root of 64!!!

\(-885y+3\)

The surface area of the prism with the isosceles right triangle base is \(685ft^2\).

The surface area of any prism can be calculated using the following formula:

Let S= Surface Area

Let L= Lateral Area (Area of everything but the base)

Let B= Area of one base

Let P= Perimeter of the base

Let H= height of prism
 

\(S=L+2B\)

\(L=PH\)

Okay, let's get started now that we have determined the formula. Let's start by solving for L, the lateral surface area.

P= perimeter of the base

\(P= (11+11+15.6)ft=37.6ft\)

\(H=15ft\)

To find the lateral surface area, multiply P by h.

\(L=PH=37.6ft*15ft=564ft^2\)

The only variable to find next is B, the area of a triangle is 1/2*bh. Because the base is a right isosceles triangle, the base and the height are both 11.

\(B=\frac{1}{2}bh=\frac{1}{2}*11*11=60.5ft^2\)

Remember, the formula requires us to find 2B:

\(2B=(2*60.5)ft^2=121ft^2\)

Now, plug L and 2B back into the formula to get the final answer:

\(S=L+2B\)

\(S=(564+121)ft^2=685ft^2\)

Normally, you would have to find the surface area of the other triangular prism, but the final answer only asked about the triangle with the right isosceles base, so there is no need to calculate the other one. 

Yes, Point C is at (-2,-2) when reflected over the x-axis and then over the y-axis.

\(9+10=19\)

Since this is not an equation, I will assume that you want to simpify this expression. This is the original:

\(7x+9*21\)

Here, the only like terms to combine is the 9*21. The rest of it cannot be simplified further.

\(7x+189\)

But you are the only one who says those things...

What you have provided is not an equation, so I will assume that you want to simplify this expression. First, this is the original equation:

\(25-5\sqrt{x}-5\sqrt{x}-x\)  

In this equation, the only like terms are the -5 multiplied by the square root of x. Therefore, the simplified expression would be:

\(25-10\sqrt{x}-x\)

the one about triangles