All Answers 356055 Answers

\((x+1)^2\\ =(x+1)(x+1)\\ = x(x+1)+(x+1)\\ =x^2+x+x+1\\ =x^2+2x+1\)

Expand the following:
(x+1)^2

(x+1) (x+1) = (x) (x) + (x) (1) + (1) (x) + (1) (1):
x x+x+x+1

x x = x^2:
x^2+x+x+1

x+x = 2 x:
Answer: |  x^2+2 x+1

84×3  =  252:
sqrt(-8252)

252 (-8)  =  -2016:
 sqrt(-2016 )

sqrt(-2016) = sqrt(-1) sqrt(2016) = i sqrt(2016):
i sqrt(2016)

sqrt(2016)  =  sqrt(32×9×7)  =  sqrt(2^5×3^2×7):
i sqrt(2^5 3^2 7)

sqrt(2^5 3^2 7)  =  sqrt(2^5) sqrt(3^2) sqrt(7)  =  2^(5/2)×3^(2/2) sqrt(7)  =  4 sqrt(2)×3 sqrt(7):
i×4 sqrt(2)×3 sqrt(7)

4×3  =  12:
i×12 sqrt(2) sqrt(7)

sqrt(2) sqrt(7) = sqrt(2×7):
i×12 sqrt(2×7)

2×7  =  14:
Answer: |  i×12  sqrt(14 )

whats the measure of Arc XY if Major Arc =110 and angle XZY=39

I will assume that Z is the centre of the circle.

first draw the pic .

360-39=131 degrees

\(\frac{341}{360}\times 2*\pi*r=110\\ \frac{341\times 2*\pi}{360}*r=110\\ r=110*\frac{360}{341\times 2*\pi}\\ \)

110*360/(341*2*pi) = 18.4825095203491358 = 18.48 units

\(Arc \;XY=\frac{39}{360}*2\pi*18.48\\\)

39/360*2*pi*18.48 = 12.5789369849735321

\(Arc\;XY\approx12.58\;units\)

 Simplify 

\(\dfrac{\log\sqrt8+\log\sqrt{125}-\log\sqrt{27}}{\log2+\log\sqrt[3]{5}-\log\sqrt[3]{12}}\)

\(\begin{array}{rcll} && \dfrac{\log\sqrt8+\log\sqrt{125}-\log\sqrt{27}}{\log2+\log\sqrt[3]{5}-\log\sqrt[3]{12}}\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\log(2)+\log{\sqrt[3]{5}} - \log{\sqrt[3]{3\cdot2^2}} }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\log(2)+\log{\sqrt[3]{5}} - \log(\sqrt[3]{3}\cdot \sqrt[3]{2^2}) }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\log(2)+\log{\sqrt[3]{5}} -\log\sqrt[3]{3} -\log\sqrt[3]{2^2} }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\log(2)+\log{\sqrt[3]{5}} -\log\sqrt[3]{3} -\log(2^\frac23 ) }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\log(2)+\log{\sqrt[3]{5}} -\log\sqrt[3]{3} -\frac23\cdot \log(2) }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\log(2)-\frac23\cdot \log(2) +\log{\sqrt[3]{5}} -\log\sqrt[3]{3} }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\frac33\cdot\log(2)-\frac23\cdot \log(2) +\log{\sqrt[3]{5}} -\log\sqrt[3]{3} }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\frac13\cdot\log(2) +\log{\sqrt[3]{5}} -\log\sqrt[3]{3} }\\\\ &=& \dfrac{\log{\sqrt{2^3}}+\log{\sqrt{5^3}}-\log{\sqrt{3^3}} }{\frac13\cdot\log(2) +\frac13\cdot\log(5) -\frac13\cdot\log(3) }\\\\ &=& \dfrac{ \frac32\cdot\log(2)+\frac32\cdot\log(5)-\frac32\cdot\log(3) }{\frac13\cdot\log(2) +\frac13\cdot\log(5) -\frac13\cdot\log(3) }\\\\ &=& \dfrac{\frac32}{\frac13} \left( \dfrac{ \log(2)+\log(5)-\log(3) }{\log(2) +\log(5) -\log(3) } \right) \\\\ &=& \dfrac{\frac32}{\frac13} \\\\ &=& \frac32 \cdot \frac31 \\\\ &=& \frac92 \\\\ &=& 4.5 \end{array}\)

I was actually trying to explain why this rule works.

Just add a zero behind if you multiply by 10.

1000 * 10 = 10000

It is long since the last person commented on this post. And I think the sub topics of mathematics are still not available yet?

I am going to suggest a new list.

List:

Algebra

Geometry

Logics

Statistics

Trigonometry

Pre-calculus

Calculus

This list is much shorter and I can't think of anything that isn't in the area of the list.

Differential Eq.s are in calculus.

Financial Maths are in Algebra due to the percentages and indices.

Surds are in Algebra.

Complex numbers are in Algebra or Geometry because there are various kinds of complex numbers question. Some involves Pythagoras Theorem.

Coordinates are in Geometry.

Pythagoras Theorem is in Geometry or Trigonometry.

I think we don't need so much categories.

10000

[A] = [A0]*e^(-k*t)

[A] = 0.7*e^(-0.0942*20)

I'll leave you to do the number crunching (you can just enter the right-hand side into the calculator on the home page here).

\(\text{I'm gonna type everything in }\LaTeX\)

\(\text{First Question:}\\ \quad \frac{1}{x}+\frac{1}{1-x}\\=\frac{1-x}{x(1-x)}+\frac{x}{x(1-x)}\\=\frac{1-x+x}{x-x^2}\\=\frac{1}{x-x^2}\)

\(\boxed{\color{red}\therefore \frac{1}{x}+\frac{1}{1-x}=\frac{1}{x-x^2}}\)

\(\text{Second Question:}\\ \quad 1+\frac{1}{x+1}\\=\frac{x+1}{x+1}+\frac{1}{x+1}\\=\frac{x+2}{x+1}\)

\(\boxed{\color{red}\therefore 1+\frac{1}{x+1}\neq \frac{2}{x+1}}\)

\(\text{Third Question:}\\\quad x+\frac{1}{x}\\=\frac{x^2}{x}+\frac{1}{x}\\=\frac{x^2+1}{x}\)

\(\boxed{\color{red}\therefore x+\frac{1}{x}=\frac{x^2+1}x}\)

\(\text{Last Question:}\\\quad\frac{1}{x-1}-\frac{1}{x+1}\\=\frac{x+1}{x^2-1}-\frac{x-1}{x^2-1}\\=\frac{x-x+1+1}{x^2-1}\\=\frac{2}{x^2-1}\)

\(\boxed{\color{red}\therefore\frac{1}{x-1}-\frac{1}{x+1}\neq \frac{2x}{1-x^2}}\)

\(\color{red}55mm:8cm\\ \color{red}= 5.5cm : 8 cm\\ \color{red}= 11cm:16cm\\\color{red}=11:16 \)

length of the round edge = \(\color{red}\pi\times 50 = 50\pi cm\)

Dont forget the straight edge.

length of the straight edge = \(\color{red} 2\times 50 = 100 cm\)

Perimeter of the semi-circle = \(\color{red} 100 + 50\pi\)

\(\dfrac{5x-1}{4}=5+\dfrac{x}{2}\\ 5x-1 = 20 + 2x\\ 3x = 21 \\ \color{red}\boxed{\color{aqua}x = 7}\)

Is 1/x+1/1-x = 1/x-x^2 ?

Nope, 1/x+1/1-x = (1-x)/(x+1)

Is 1+ 1/x+1 = 2/x+1 ?

Nope, 1 + 1/(x+1) = (x+1)/(x+1) + 1/(x+1) = (x+2)/(x+1)

Is x+ 1/x = x^2+1/x ?

Yes, x + 1/x = x^2/x + 1/x = (x^2+1)/x

Is 1/x-1 - 1/x+1 = 2x/1-x^2 ?

Nope, 1/(x-1) - 1/(x+1) = (x+1)/(x-1) - (x-1)/(x+1) = ((x+1)-(x-1))/(x^2-1) = 2/(x^2-1)

A helpful way to check your answer is to graph the equation (that way you can instantly visually see if they equal each other). 

Let f(x) = 4x-5 and g(x) = 3x, find (f/g)(x).

I think .......

\((f/g)(x)=\frac{4x-5}{3x}\)

Calculate the simple interest earned on an investment of $1050 at a semiannual rate of $1.1 for $9 years. Write your answer to the nearest cent.

Your question does not make sense.  Not to me anyway.

Have a look at:  http://www.piday.org/million/ 

Thanks Chris, I forgot that shortcut :)

Thanks Max,

In any circle pi = circumference divided by the diameter.  

The ratio (fraction) always is the same. :)

It is a right angle triangle so use the pythagorean theorum :)

find number of distinguishable permutations of 2 E's, 3 J's and 4 L's

\(\frac{9!}{2!*3!*4!}= \frac{5*6*7*8*9}{2*2*3}=\frac{5*7*4*9}{1}=5*2*2*7*9=126*10=1260\)

Mmm, Some of your answers I agree with Max but some are open to interpretation.

Which of these functions is a linear function?

a. f(x) = 2x²     yes defintely NOT a lnear graph.

 Linear graphs cannot have x raised to any power (except 1 which is invisable) 

b. f(x) = (1 / 2)²x  =   \((\frac{1}{2})^2x = \frac{1}{4}x=0.25x \qquad \mbox{YES this is definitely a LINEAR equation}\)

c. f(x) = 1 / 2x 

If you mean f(x)=1/(2x) then it is not linear becasue x cannot be on the bottom.

If you mean  (1/2)*x then it IS LINEAR

d. f(x) = - 1 / 2(x - 2)

If you mean\( f(x)=\frac{-1}{2(x-2)}\)

                       then it is NOT linear becasue x cannot be on the bottom.

If you mean  \( f(x)=\frac{-1}{2}(x-2)\)

                     then it IS LINEAR