Melody

Thanks Alan,

Usin Alan's pic

(II) if SinB+SinC=1, then what is the shape of this triangle?

sinB = 1/2

SinC=1/2

so statement is true

So B=C=pi/6    is one answer.

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May I ask 2 questions please Alan

1) How do you know that is the only answer?    and

2) Why didn't Wolfram|alpha give me that answer?

Thankyou.:)

$$\\f(x)= 3x^3-x^2-\frac{2x}{x}(x+1)^2\\
x\ne0\\
f(x)= 3x^3-x^2-2(x+1)^2\\$$

A team consists of 6 boys and 4 girls and the other has 5 boys and 3 girls. How many single matches can be arranged between the two teams when a boy plays against a boy and a girl plays against a girl?

Consider team A

girl 1 three to choose from

girl 2 three to choose from

etc

3*4=12

boys

6*5=30

Total=  12+30=42

Maybe, if you are very lucky that might be right.  

I am probably interpreting the question incorrectly BUT i don't think the second parabola is relevant.

Let the vertex of the concave up parabola be (0,0)

The 2 top points are (-75,70) and (75,70)

The equation is

$$\\x^2=4ay\\
75^2=4a*70\\
5625=280a\\
a\approx20.09\\$$

so

$$\\x^2\approx 4*20.09y\\
x^2\approx 80.36y\\
$Find x when y=35\\
x^2\approx 80.36*35\\
x^2\approx 2812.5\\
x\approx 53$$

Maybe I didn't understand where the 75 belonged  

YOU CALLED ?

THIS HAD BETTER BE IMPORTANT

I TRAVELLED A LOOOONG WAY (THROUGH TIME) TO BE HERE

ROARRRRRRRRRRRRRRR

Part 2

(II) if SinB+SinC=1, then what is the shape of this triangle?

$$\\ SinB+SinC=1\\
Sin\left(\frac{\pi}{3}-C\right)+SinC=1\\$$

Now I am really confused because according to Wolfram|Alpha there is no valid solution to this.  Where B and C are acute angles. 

In a triangle ABC, a, b and c are the opposite sides of each angles ,and 2a*SinA=(2b+c)*SinB+(2c+b)*SinC

(I） try to the angle A

$$\\2aSinA=(2b+c)SinB+(2c+b)SinC\\
0=(2b+c)SinB+(2c+b)SinC-2aSinA\\$$

Using the sine rule I know that

$$\\SinB=\frac{bSinA}{a}\qquadand\qquad SinC=\frac{cSinA}{a}\\\\
$Substituting I get$\\\\$$

$$\\0=(2b+C)\frac{bSinA}{a}+(2c+b)\frac{cSinA}{a}-2aSinA\\\\
0=SinA[(2b+C)\frac{b}{a}+(2c+b)\frac{c}{a}-2a]\\\\
0=\frac{SinA}{a}[(2b+c)b+(2c+b)c-2a^2]\\\\
$Neither sinA nor A can be 0 so$\\\\$$

$$\\0=[2b^2+cb+2c^2+bc-2a^2]\\\\
0=[2b^2+2c^2-2a^2+2bc]\\\\
0=b^2+c^2-a^2+bc\\\\
-bc=b^2+c^2-a^2\\\\
-\frac{1}{2}=\frac{b^2+c^2-a^2}{2bc}\\\\
-\frac{1}{2}=cosA\qquad\qquad \mbox{Using cosine rule}\\\\
A=\pi-\frac{\pi}{3}\\\\
A=\frac{2\pi}{3}\\\\$$

That is true but

5x+6x=11x

There are 10 grams of Curium-245 which has a half-life of 9300 years. How many grams will remain after 37200 years?

$$\begin{array}{rll}
A&=&A_0e^{kt}\\\\
M&=&2Me^{k*9300}\\\\
0.5&=&e^{k*9300}\\\\
ln(0.5)&=&ln(e^{k*9300})\\\\
ln(0.5)&=&(k*9300)ln(e)\\\\
ln(0.5)&=&k*9300\\\\
k&=&\frac{ln0.5}{9300}\\\\
So&&\\\\
A&=&A_0e^{\frac{ln0.5}{9300}*t}\\\\
&&$Start with 10g. How many grams will remain after 37200 years?$\\\\
A&=&10e^{\frac{ln0.5}{9300}*37200}\\\\
\end{array}$$

$${\mathtt{10}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{\left({\frac{{{log}}_{{\mathtt{e}}}{\left({\mathtt{0.5}}\right)}}{{\mathtt{9\,300}}}}{\mathtt{\,\times\,}}{\mathtt{37\,200}}\right)} = {\frac{{\mathtt{5}}}{{\mathtt{8}}}} = {\mathtt{0.625\: \!000\: \!000\: \!000\: \!000\: \!1}}$$

so        625mg will remain