Melody

It is a fraction button EP..

It does make fractions much easier to deal with.

But it is certainly not essential for anyone who has a good grasp of primary school maths.

It is not there because the designer of the calculator considered it unnecessary.

I am taking this straight from EPs answer.  I have not looked if EP's answer is correct

\(y = \frac{2x^2}{ 3} - 4x + 4 \\ y -4= \frac{2x^2}{ 3} - 4x \\ \frac{3}{2}(y -4)= \frac{3}{2}( \frac{2x^2}{ 3} - 4x) \\ \frac{3}{2}(y -4)= x^2 - 6x \\ \frac{3}{2}(y -4)+3^2= x^2 - 6x +3^2 \\ \frac{3}{2}(y -4)+9= x^2 - 6x +9 \\ \frac{3}{2}(y -4)+9= (x-3)^2 \\ \frac{2}{3}(\frac{3}{2}(y -4)+9)= \frac{2}{3} (x-3)^2 \\ y -4+6= \frac{2}{3} (x-3)^2 \\ y+2= \frac{2}{3} (x-3)^2 \\ \frac{3}{2}(y+2)= (x-3)^2 \\ (x-3)^2=4*\frac{3}{8}(y+2)\)

This is a concave up parabola with vertex = (3,-2)   and focal length a=3/8

I have not checked my answer.

This is very simple.

All you have to know is  tanx = sinx / cosx

(e^(x^3/6))/(e^(x^2/2))

(e^(x^3   /6))   /   (e^(x^2    /2))

\(\frac{e^{\frac{x^3}{6}}} {e^{\frac{x^2}{2}}}\\~\\ =\frac{e^{\frac{x^3}{6}}} {e^{\frac{3x^2}{6}}}\\~\\ =e^{\frac{x^3-3x^2}{6}} \\~\\ \)

A circle is inscribed in a right isosceles triangle whose legs are 8 inches long. How many square inches are in the area of the shaded portion? Express your answer as a decimal rounded to the nearest hundredth.

Think of it as a co-ordinate geometry question

the vertices are   (0,0)   (0,8)   (8,0)

Let (r,r) be the centre of the circle

x=0,  y=0  and  y=-x+8  are all tangents to the circle

Points on the circumference are  (r,0), (0,r) (4,4)

Distance from the centre to (0,0) = \(\sqrt{2r^2}=\sqrt2\;r\)

Distance from origin to (4,4) =\( \sqrt{32}=4\sqrt2\)

Distance from origin to (4,4) in terms of raius r is  \( \sqrt2 \;r+r=r(\sqrt2+1)\)

So

\(r(\sqrt2+1)=4\sqrt2\\ r=\frac{4\sqrt2}{\sqrt2+1}=\frac{4\sqrt2(\sqrt2-1)}{1}=8-4\sqrt2=4(2-\sqrt2)\)

Area of circle = \(\pi( 4(2-\sqrt2))^2=16\pi(4+2-4\sqrt2)=32\pi(3-2\sqrt2)\)

Area of triangle = 0.5*8*8 = 32

\(\text{Shaded area}=32-32\pi(3-2\sqrt2)\\ \text{Shaded area}=32(1-3\pi+2\pi\sqrt2)\\ \text{Shaded area}\approx 14.75\; square\; inches\)

You need to check my working.

It is an AP

5,10, .....  395  

395/5 = 79

sum=n/2(a+L) = 79/2  *  (5+395) = 79/2  * 400 = 79*200 = 15800

I have not looked much but the angles will all be 120 degrees.

Use cosine rule to find distance AE

triangle AFE and triangle DBC are congruent

so AE = BD

Area of AFE can be determined.

ABDE is a parallelogram. I am pretty sure it is a rectangle but i have not looked at it enough to be sure.

If it is a rectangle then the area is easily obtained.

That is a rough outline of some of the steps you will need to use.

Yes you do have to use Vieta's formulas.

First you have to get a common denominator.

I have drawn it to scale and have the answer 10 degrees but i cannot see why.

Thanks Geno, that is a excellent method. 

It always pays to do a quick sketch and then facts like this are more likely to be seen.

Here is the super long methods.

I'd  have let the centre of the circle be (x,y)

Then found the distances and said they had  to be equal.

\(r^2=(x-6)^2+(y-2)^2\qquad(1)\\ r^2=(x-8)^2+(y-6)^2\qquad (2)\\ r^2=(x-8)^2+(y-10)^2 \qquad (3)\\~\\ (3)-(2)\\ 0=(y-6)^2-(y-10)^2\\ 0=y^2-12y+36-(y^2-20y+100)\\ 0=y^2-12y+36-y^2+20y-100\\ 0=8y-64\\ 0=y-8\\ y=8\\~\\ sub\;\; in\\ r^2=(x-6)^2+(6)^2\qquad(1b)\\ r^2=(x-8)^2+(2)^2\qquad (2b)\\ subtract\\ 0=(x-6)^2+36-(x-8)^2-4\\ 0=(x^2-12x+36)+36-(x^2-16x+64)-4\\ 0=-12x+36+36+16x-64-4\\ 0=4x+4\\ x=-1 \)

\(r^2=(x-2)^2+(y-6)^2\qquad(1)\\ r^2=(x-8)^2+(y-6)^2\qquad (2)\\ r^2=(x-8)^2+(y-10)^2 \qquad (3)\\~\\ (3)-(2)\\ 0=(y-6)^2-(y-10)^2\\ 0=y^2-12y+36-(y^2-20y+100)\\ 0=y^2-12y+36-y^2+20y-100\\ 0=8y-64\\ 0=y-8\\ y=8\\~\\ sub\;\; in\\ r^2=(x-2)^2+(0)^2\qquad(1b)\\ r^2=(x-8)^2+(0)^2\qquad (2b)\\ subtract\\ 0=(x-2)^2-(x-8)^2\\ 0=(x^2-4x+4)-(x^2-16x+64)\\ 0=-4x+4+16x-64\\ 0=12x-60\\ x=5\\~\\ \text{So the centre is (5,8)} \)

Coding:

r^2=(x-2)^2+(y-6)^2\qquad(1)\\
r^2=(x-8)^2+(y-6)^2\qquad (2)\\
r^2=(x-8)^2+(y-10)^2 \qquad (3)\\~\\

(3)-(2)\\
0=(y-6)^2-(y-10)^2\\
0=y^2-12y+36-(y^2-20y+100)\\
0=y^2-12y+36-y^2+20y-100\\
0=8y-64\\
0=y-8\\
y=8\\~\\
sub\;\; in\\
r^2=(x-2)^2+(0)^2\qquad(1b)\\
r^2=(x-8)^2+(0)^2\qquad (2b)\\
subtract\\
0=(x-2)^2-(x-8)^2\\
0=(x^2-4x+4)-(x^2-16x+64)\\
0=-4x+4+16x-64\\
0=12x-60\\
x=5\\~\\
\text{So the centre is (5,8)}