Rom

How can I remake function from y= log(x) to x=smth y ? 

 

I've got function y = 134.94ln(x) + 991.04 and I need to make the other form (x=something) 

\(x = e^{\ln(x)}\\ \\ e \text{ is the base of the natural logarithms. }e \approx 2.718\) 

\(y=a \ln(x) + b \\ \ln(x) = \dfrac{y-b}{a}\\ \\ x = e^{ \frac{y-b}{a}}\)

and in your case

\(\Large x = e^{\frac{y - 991.04}{134.94}}\)

Well, infinity plus infinity(if you were associating infinity as a number) would make a bigger number altogether,

No.

The set of symbols

\(\infty + \infty > \infty\) 

is meaningless.  You cannot do arithmetic with infinity.

\(\dfrac 1 0 \text{ is undefined.}\)

\(\infty - \infty \text{ is meaningless, you can manipulate this to be any value you like.}\\ \\ \text{It certainly doesn't equal 0.}\)

\(\text{The whole "proof" is nonsense.}\)​

Two workers, if they were working together, could finish a certain job in 12 days. If one of the workers does the first half of the job and then the other one does the second half, the job will take them 25 days. How long would it take each worker to do the entire job by himself?

The key to this is that the rates they work at add.

\(\text{We get 3 equations.}\\ \\ (r_1+r_2)12=1 \\ \\ r_1 t = r_2(25-t) =\dfrac 1 2\\ \\ r_1 t + r_2 (25-t)=1\)

you can pound through these equations to obtain

\(r_1=\dfrac 1 {20},~~r_2=\dfrac{1}{10}\\ \\ \text{And thus worker 1 would finish in }\dfrac 1{r_1} = 20 \text{ days.}\\ \\ \text{and worker 2 would finish in }\dfrac{1}{r_2} = 10 \text{ days.}\) 

In a vector space, how much null vectors exist?

\(\text{The most common use of the term null vector refers to the vector consisting of all 0's.}\\ \text{Thus in a given vector space there exists only one of these.}\\ \\ \text{A less common use refers to a vector }x \text{, such that for a given matrix }A, A x = 0\\ \\ \text{If there is such a vector other than the vector of all 0's,}\\ \text{ then there are infinity of them as any scalar multiple of }x\\ \\ \text{will also satisfy the definition.}\)

4a square -160+12a

you all usually want to know the roots of these quadratics when you just toss em up so

\(4a^2 +12 a - 160=0 \\ \\ a^2+ 3a - 40 = 0 \\ \\ (a+8)(a-5)=0 \\ \\ a = -8 \text{ or }a=5\) 

the top is 12x15 =180 sq in

two sides are 4x12=48 sq in so 96 sq in total

two sides are 4x15=60 sq in so 120 sq in total

180 + 96 + 120 = 396 sq in of frosting area

are you allowing leading 0's or not?

If not then in order for one of these numbers to be divisible by 45 it must end, and thus start, with 5.

I show the largest is 59895, and the smallest is 50805, and the difference is 9090.

If you are including numbers such as 01510, the answer will be different.

The surface area of a hexoganal pyramid with lateral edges of 13cm and base edges of 10cm?

You have 6 triangles that are 13x13x10cm

And you have the regular hexagon pyramid base with each side length 10cm.

The area of each of the triangles is given by

\(A_{face} = b h = b \sqrt{s^2-\dfrac{b^2}{4}}cm^2\)

there are 6 of these so the total area of the faces is 

\(A_{sides}=6b\sqrt{s^2-\dfrac {b^2}{4}}cm^2\)

Now the area of the hexagonal base is given by

\(A_{base}=\dfrac{3\sqrt{3}}{2}b^2cm^2\)

plugging the values for b and s in we get

\(A = A_{sides}+A_{base} = 6(10)\sqrt{(13)^2-\dfrac{(10)^2}{4}}+\dfrac{3\sqrt{3}}{2}(10)^2=\\ \\ 720+150 \sqrt{3}cm^2\)

Suppose you have a closed shape.  It can be anything as long as it's closed.

Now pick a point and start walking around the border of this shape.  Eventually, because it is closed, you will reach the point you started from.

The total distance you travelled along the border of the shape is called it's perimeter.

Similarly imagine snipping the border of the shape in two at some point and drawing it out into a straight line without stretching it.  The length of this line is the shape's perimeter.