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\(2x^2-(x+3)-2x(1-x)\\ =2x^2-x-3-2x+2x^2\\ =4x^2-3x-3\)

\((2xy^2)^3\div(3x^2y^3)\times (9x^3y^2)\\ =(2^3\div 3\times 9)\times (x^3\div x^2 \times x^3)\times ((y^2)^3\div y^3\times y^2)\\ =24x^4y^5\)

\(\dfrac{5}{6}\div \dfrac{3}{6}\times \dfrac{3}{15}\\ =\dfrac{5}{6}\div \dfrac{1}{2}\times \dfrac{1}{5}\\ =\dfrac{5}{6}\times 2 \times \dfrac{1}{5}\\ =\dfrac{5\times 2}{6\times 5}\\ =\dfrac{10}{30}\\ =\dfrac{1}{3}\)

1/8 = 0.125, so we have 3 1/8 = 3 + 1/8 = 3 + 0.125 = 3.125.

Do you mean partial fractions?

\(\dfrac{1}{x^2 - 1}=\dfrac{a}{x-1}+\dfrac{b}{x+1}\\ 1=a(x+1)+b(x-1)=(a+b)x+a-b \\a=1/2\text{ and }b=-1/2\\ \therefore \dfrac{1}{x^2-1}=\dfrac{1}{2(x-1)}-\dfrac{1}{2(x+1)}\)

We made 1/(x^2-1) seperate into 2 fractions: 1/(2(x-1)) and -1/(2(x+1)).

\(\text{LCM}(9a^2bc,8abc,6ab^3)\\ =\text{LCM}(9,8,6)\cdot a^2b^3c\\ =72a^2b^3c\)

\(\text{LCM}(xy,x^2,xy^3)\\ \text{You find the highest power of x and y among the 3 and multiply together.}\\ =x^2y^3\)

"Triangle PQR is right angled at Q, and PQ is 6cm long. QR increases in length at 2cm per minute. Find the rate of change in QPR at the instant when QR is 8cm Long."

.

Divide and substitute and you should get the following:

w = 2 1/2,  x = 6,  y = 2/3,  z = 3, so have:

w+x+y+z = 2 1/2 + 6 + 2/3 + 3 =12 1/6

3.125?

You don't have the denominator (i.e. the number under the line in between them.)

Input:2 x 2 - ( x + 3) - 2 x (1 - x)

Intepretation:\(2x^2-(x+3)-2x(1-x)\)

\(=2x^2-x-3-2x+2x^2\)

\(=4x^2-3x-3\)

Q.E.D.

Input:xy+ 3x2y+ 1 + 2xy2+yx2+ 2 +yx

Intepretation:\(xy+3x^2y+1+2xy^2+yx^2+2+yx\)

Simplify:

\(=3+2xy+4x^2y+2xy^2\)

(Function non-further-factorizatable)

(If you wanted us to do four questions, please put them all in one question rather than serving all of them in quick succession.)

Input:( 2 xy 2 ) 3 ÷ 3 x 2 y 3 × 9 x 3 y 2

Intepretation: \(\frac{\left(2xy^2\right)^3}{3x^2y^3}\cdot 9x^3y^2\)

Simplification:

\(=\frac{\left(2xy^2\right)^3}{y}\cdot 3x\)

Extract 2 out of the parenthese:

\(=\frac{8\left(xy^2\right)^3}{y}\cdot 3x\)

Continue Simplification:

\(=\frac{8(x^3y^6)}{y}\cdot 3x\)

\(=8x^3y^5\cdot 3x\)

\(=24x^4y^5\)

Q.E.D.

expression is x/p

differentiate (2x+7) (chain rule) than you get denominator equal 2

\(12 {1\over2}\)

Stewart cycles at an average speed of 10 kilometres per hour.

How many days will it take him to travel 1000 kilometres,

if he cannot cycle more than 8 hours per day?

\(\begin{array}{|rcll|} \hline && 1000\ km \cdot \frac{1\ \text{hour}}{10\ km}\cdot \frac{1\ \text{day}}{8\ \text{hour}} \\ &=& \frac{1000}{10\cdot8} \ \text{days} \\ &=& \frac{100}{8} \ \text{days} \\ &=& \frac{25}{2} \ \text{days} \\ &=& 12.5 \ \text{days} \\ \hline \end{array}\)

It will take him to travel 1000 kilometers about 12.5 days

For this problem, we have to divide 1000 km by the product of hours per day and average speed:

\(days=\frac{1000}{10*8=80}\)

\(days=\frac{960+40}{80}\)

\(days=\frac{960}{80}+\frac{40}{80}\)

\(days=12+\frac{1}{2}\)

\(days = 12.5\)

Stewart will take 12.5 days to travel 1000km.

Interesting enough... No wonder it burnt your brain, here we go:

Illustration (Made om quick session in 3 mins)

We know the height of the building is 33 meters, therefore we obtain the following equation:

Set the distance from the foundation of the building to the church's foundation \(A\).\

Apply trigonometric identity.

The angle of  depression is 25 degrees, therefore:

\(A=tan(25°)*33\)

\(A≈ 0.4663*25\)

\(A≈ 11.6575\)

That means the distance from the church's top to the same height of the building is also \(A=11.6575\)

Applying trigonometric identites for the second time

Church height=\(11.6575*sec(18°)\)

\(≈11.6575*1.0515\)

\(=12.258\)=Church's Height

Q.E.D.

(Don't even ask me about how tensed I am to win over Heureka over the speed of solving this question)

A person on top of a building 33 metres high looks across the road to a church on the other side.

The angle of depression of the bottom of the church is 25 degrees and

the angle of elevation of the top of the church is 18 degrees.

Calculate the height of the church.

Let h = height of the church
Let d = horiz. distance building and church

\(\begin{array}{|rcll|} \hline (1) & \tan(18^{\circ}) &=& \frac{h-33\ m}{d} \\ (2) & \tan(25^{\circ}) &=& \frac{33\ m}{d} \\ \hline & d = \frac{h-33\ m}{\tan(18^{\circ})} &=& \frac{33\ m}{\tan(25^{\circ})} \\ & \frac{h-33}{\tan(18^{\circ})} &=& \frac{33}{\tan(25^{\circ})} \\ & h-33 &=& \frac{ \tan(18^{\circ}) } { \tan(25^{\circ}) } \cdot 33 \\ & h &=& 33 + \frac{ \tan(18^{\circ}) } { \tan(25^{\circ}) } \cdot 33 \\ & h &=& 33\cdot \left(1 + \frac{ \tan(18^{\circ}) } { \tan(25^{\circ}) } \right) \\ & h &=& 33\cdot \left(1 + \frac{ 0.32491969623 } { 0.46630765815 } \right) \\ & h &=& 33\cdot 1.69679253718 \\ & h &=& 55.9941537270 \ m\\ \hline \end{array}\)

The height of the church is \(\mathbf{\approx 56\ m}\)

For you to calculate the answer via this calculator, type in the following code:

(-6)/(-2)

The fraction symbol is /.

The answer would be:

\(\frac{\left(-6\right)}{\left(-2\right)}=3\)

Q.E.D.

\(\frac{27}{64}=0.421875\)

\(\frac{27}{32}=0.84375\)