fiora

the given in question one is not vaild, when you doing number two.

see my link on demos,https://www.desmos.com/calculator/a5dnftk4qi

No.Here is a graph from desmos:

Hey,Melody!

I had mention that remenber "If" at the end of my question.That mean the given in question one is not vaild, when you doing number two.

I guess a few picture can help you understand my question.

Angle NOM is moving around in triangle QOP,what is the possible mininum area of triangle NOM?and find the measure of angle MOP.Sorry Melody.

you got half of the anwsers,try it in another way,like using law of cosine

Awesomeee's question is 21^2+22^2+...+40^2

not 1^2+2^2+3^2+4^2+....39^2+40^2

21^2+22^2+...+40^2=[(1^2+2^2+3^2+...39^2+40^2)-(1^2+2^2+...19^2+20^2)]/6

                                =$${\frac{\left[{\mathtt{40}}{\mathtt{\,\times\,}}\left({\mathtt{40}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right){\mathtt{\,\times\,}}\left({\mathtt{81}}\right){\mathtt{\,-\,}}{\mathtt{20}}{\mathtt{\,\times\,}}\left({\mathtt{20}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right){\mathtt{\,\times\,}}{\mathtt{41}}\right]}{{\mathtt{6}}}} = {\mathtt{19\,720}}$$

12.5cm Squared times 100mm2 over 1cm2?

1cm=10mm

(1cm)^2=(10mm)^2

$${\mathtt{1}}\left[{{cm}}^{{\mathtt{2}}}\right] = {\mathtt{100}}\left[{{mm}}^{{\mathtt{2}}}\right]$$

I guess you mean 12.5cm Squared is (12.5cm)^2,so the question could represnt as:

$${\frac{{\left({\mathtt{12.5}}{cm}\right)}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{100}}\left[{{mm}}^{{\mathtt{2}}}\right]}{{\mathtt{1}}\left[{{cm}}^{{\mathtt{2}}}\right]}} = {\left({\frac{{\mathtt{25}}}{{\mathtt{2}}{cm}}}\right)}^{{\mathtt{2}}} = {\frac{{\mathtt{625}}}{{\mathtt{4}}}}{\mathtt{\,\times\,}}{{\mathtt{cm}}}^{{\mathtt{2}}} = {\mathtt{156.25}}\left[{{cm}}^{{\mathtt{2}}}\right]$$

Thank you,Cphill! Your anwser is correct.see,.

Thank you so much for taking a time to solve it and explain to me.

$${{\mathtt{x}}}^{{\mathtt{2}}} = {\mathtt{400}}$$

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{400}} = {\mathtt{0}}$$

$$\left(\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{20}}\right){\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{20}}\right)\right) = {\mathtt{0}}$$

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{x}}={\mathtt{20}}\\
{\mathtt{x}}=-{\mathtt{20}}\end{array}\right)}$$

Thank you,heureka!

I noticed that I could find the measure of angle ADC in this way.

angle ADC +angle DAC +angle ACD =180 degrees

angle ADC=180-(angle DAC+angle ACD)

cosADC=cos[180-(angle DAC+angle ACD)]

cosADC=$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{\pi}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{DAC}}{\mathtt{\,\small\textbf+\,}}{\mathtt{ACD}}\right)}{\mathtt{\,\small\textbf+\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{\pi}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{DAC}}{\mathtt{\,\small\textbf+\,}}{\mathtt{ACD}}\right)}$$

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{\pi}}\right)} = {\mathtt{0}}$$

therefore,cosADC=$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{\pi}}\right)}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{DAC}}{\mathtt{\,\small\textbf+\,}}{\mathtt{ACD}}\right)}$$=$${\mathtt{\,-\,}}\left({\mathtt{1}}{\mathtt{\,\times\,}}\left({\mathtt{cosDAC}}{\mathtt{\,\times\,}}{\mathtt{cosACD}}{\mathtt{\,-\,}}{\mathtt{sinDAC}}{\mathtt{\,\times\,}}{\mathtt{sinACD}}\right)\right) = {\mathtt{sinDAC}}{\mathtt{\,\times\,}}{\mathtt{sinACD}}{\mathtt{\,-\,}}{\mathtt{cosDAC}}{\mathtt{\,\times\,}}{\mathtt{cosACD}}$$

given that $$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{cosdac, cosacd}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{cosDAC}}={\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}\\
{\mathtt{cosACD}}={\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}}{{\mathtt{5}}}}\end{array}\right)}$$

so $$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{sindac, sinacd}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{sinDAC}}={\frac{{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}\\
{\mathtt{sinACD}}={\frac{{\sqrt{{\mathtt{5}}}}}{{\mathtt{5}}}}\end{array}\right)}$$

cosADC=sinDAC*sinACD-cosDAC*cosACD=$${\frac{{\frac{{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}}{{\mathtt{5}}}}{\mathtt{\,-\,}}{\frac{{\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{5}}}}}{{\mathtt{5}}}} = {\mathtt{\,-\,}}{\frac{{\mathtt{5}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{50}}}}}{{\mathtt{50}}}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{50}}}}}{{\mathtt{10}}}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}$$

angle ACD=arccos -sqrt(2)/2=135 (the measure of angle ACD is less than 180 degrees,ut more than 90 degrees)

AC=$${\mathtt{AC}} = {\frac{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{135}}^\circ\right)}}{\left[{\frac{{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}\right]}} = {\frac{\left[{\frac{{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}\right]}{\left[{\frac{{\sqrt{{\mathtt{10}}}}}{{\mathtt{10}}}}\right]}} = {\frac{{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{10}}}} = {\sqrt{{\mathtt{5}}}}$$ 

AC+BC=$${\sqrt{{\mathtt{5}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}$$

60 cents =0.6 dollar

85*0.6+15.50=51+15.5=66.5 dollar