Dimitristhym

What is the symbol $ ? 

Why is 15$ and not 15.What is the difference? 

Thank you.

No

area doghouse = (area triangle) + (area square)

area doghouse = (\(\frac{1}{2}\times\)base \(\times\)height ) + (side)^2

area doghouse = (\(\frac{1}{2}\times3\times5\)) + \((5)^2\)

area doghouse = \((\frac{15}{2})\) + \((25)\)

area doghouse = \(7.5+25\)

area doghouse = \(32.5\) square feet

Corect answer the 1st.If you don't understand something tell me or you can send me in message.

I hope I help you.

The triangle right as you say: \(\frac{3\times5}{2}=7.5\)

How you find the area of square? Do you remember the type? 

Nickolas because I see you upload a lot of cuestions but there are similar.When someone solve it,try to understand it and try to solve the other.

I like your advices CPhill.I like too to do a lesson with questions to the student.That's a friendly advice.

Have a good time! 

Yes i calculate 150 rads :p

My bad I will edit it.

It's false the pronunciation we have 8 different letters and at first FIVE-FOUR=ONE  we have:

E-R=E <=> R=0 But you say  different digits from one to ten, 1-10. 

1) We will use Law of cosines:

\((CB)^2=(CA)^2+(BA)^2-2(CA)(BA)cosA<=>2^2=3^2+4^2-(2\times3\times4cosA)<=>-21=-24cosA<=>cosA=\frac{7}{8}\)

So \(cosA=\frac{7}{8}\) or \(cosA=0.875\)

2)We will do pythagorean theorem at ACD : \((AC)^2= (DC)^2 + (AC)^2<=> 10^2=6^2 + (AC)^2<=>(AC)^2=10^2-6^2<=>AC=\sqrt{100-36}=\sqrt{64}=8\)

So \(AC=8 \)

\(sinB=\frac{AC}{BC}=\frac{8}{15}\) or \(sinB=0.533...\)

3) We will use Law of cosines (like 1):

\((BC)^2=(AB)^2+(AC)^2-2(AB)(AC)cos150<=>174=(CA)^2+81+15.588CA<=>(CA)^2+15.588CA-66=0\)

CA=x => \(x^2+15.588x-66=0<=>x=\frac{-15.588±\sqrt{506.985744}}{2}\)

\(x1=3.4641...\) , \(x2=-19.0521...\)

x2<0 We want distance witch always > 0 

So  \( AC=x1=3.4641...\)

BUT It's more easy to solve it like CPhill solution with roots!

This is a different way to solve it.

Hope I help you!

Oh I understand it! 

Thank you very much.The first \(τ\) it's wrong.

Because it's  \(32!/((32-5)!*5!)\) you need a () at the denominator and the practice multiplier * between ! and 5!

\(cos^2x+sin^2x=1<=> h^2 + sin^212=1<=>sin12= \sqrt{1-h^2}=\sqrt{(1-h)(1+h)}\)

\(cot12=\frac{cos12}{sin12}=\frac{h}{\sqrt{(1-h)(1+h)}}\)