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8)
Find DB in the diagram below.
Picture: https://latex.artofproblemsolving.com/f/6/f/f6f917a5c86b363a5cdf8871300d061dcd54fc0b.png

\(\begin{array}{|rcll|} \hline \angle C &=& 180^{\circ} - ( 30^{\circ} + 45^{\circ}) \\ \angle C &=& 105^{\circ} \\\\ \angle ACD &=& \angle C - 60^{\circ} \\ \angle ACD &=& 105^{\circ} - 60^{\circ} \\ \angle ACD &=& 45^{\circ} \\\\ \dfrac{DC} {\sin(30^{\circ})} &=& \dfrac{2}{\sin(ACD)} \\\\ \dfrac{DC} {\sin(30^{\circ})} &=& \dfrac{2}{\sin(45^{\circ}) } \\\\ DC &=& \dfrac{2\cdot \sin(30^{\circ}) }{\sin(45^{\circ}) } \quad & | \quad \sin(30^{\circ}) = \dfrac{1}{2} \\\\ DC &=& \dfrac{2\cdot \dfrac{1}{2} }{\sin(45^{\circ}) } \\\\ DC &=& \dfrac{1}{\sin(45^{\circ}) } \quad & | \quad \sin(45^{\circ}) = \dfrac{\sqrt{2}}{2} \\\\ DC &=& \dfrac{1}{\dfrac{\sqrt{2}}{2} } \\\\ DC &=& \dfrac{2}{ \sqrt{2} }\cdot \dfrac{\sqrt{2}}{\sqrt{2}} \\\\ DC &=& \dfrac{2\sqrt{2}}{2 } \\\\ \mathbf{DC} & \mathbf{=} & \mathbf{ \sqrt{2} } \\\\ \dfrac{DB} {\sin(60^{\circ})} &=& \dfrac{\sqrt{2}}{\sin(45^{\circ})} \\\\ DB &=& \dfrac{\sqrt{2}\cdot \sin(60^{\circ})}{\sin(45^{\circ})} \quad & | \quad \sin(45^{\circ}) = \dfrac{\sqrt{2}}{2} \\\\ DB &=& \dfrac{\sqrt{2}\cdot \sin(60^{\circ})}{\dfrac{\sqrt{2}}{2}} \\\\ DB &=& 2\cdot \sin(60^{\circ}) \quad & | \quad \sin(60^{\circ}) = \dfrac{\sqrt{3}}{2} \\\\ DB &=& 2\cdot \dfrac{\sqrt{3}}{2} \\\\ \mathbf{DB} & \mathbf{=} & \mathbf{ \sqrt{3} } \\ \hline \end{array}\)

7)
In triangle ABC, sin A = 4/5. Find cos B.
Picture: https://latex.artofproblemsolving.com/c/8/3/c83240e209d830cd2275ed9f771185a1d1c17733.png

sin-rule

\(\begin{array}{|rcll|} \hline \dfrac{\sin(B)} {10} &=& \dfrac{\sin(A)}{17} \\\\ \sin(B) &=& \dfrac{10\cdot \sin(A)}{17} \quad & | \quad \sin(A) = \dfrac {4}{5} \\\\ \sin(B) &=& \dfrac{10\cdot \dfrac {4}{5}}{17} \\\\ \sin(B) &=& \dfrac{2\cdot 4 }{17} \\\\ \mathbf{\sin(B)} & \mathbf{=} & \mathbf{ \dfrac{8}{17} } \\\\ \cos(B) &=& \sqrt{1-\sin^2(B)} \\ \\ \cos(B) &=& \sqrt{1- \left(\dfrac{8}{17} \right)^2 } \\ \\ \cos(B) &=& \sqrt{1- \dfrac{8^2}{17^2} } \\ \\ \cos(B) &=& \dfrac{ \sqrt{17^2-8^2} } {17} \\ \\ \cos(B) &=& \dfrac{ \sqrt{225} } {17} \\ \\ \mathbf{\cos(B)} & \mathbf{=} & \mathbf{ \dfrac{ 15 } {17} } \\ \hline \end{array}\)

6)

Find cos A.
Picture: https://latex.artofproblemsolving.com/6/4/a/64ae47768bae6b24ab792c44a92c5de244955cad.png

cos-rule

\(\begin{array}{|rcll|} \hline 2^2 &=& 3^2+4^2-2\cdot 3 \cdot 4\cdot \cos(A) \\ 4 &=& 9+ 16 -24\cdot \cos(A) \\ 4 &=& 25 -24\cdot \cos(A) \quad & | \quad +24\cdot \cos(A) \\ 4 +24\cdot \cos(A) &=& 25 \quad & | \quad -4 \\ 24\cdot \cos(A) &=& 25 -4 \\ 24\cdot \cos(A) &=& 21 \quad & | \quad : 24 \\ \cos(A) &=& \dfrac{21} {24} \\\\ \mathbf{\cos(A)} & \mathbf{=} & \mathbf{ \dfrac{7} {8} } \\ \hline \end{array}\)

5)
Find sin B.
Picture:https://latex.artofproblemsolving.com/4/7/a/47a312412eff41c09e4ce21e5edbb7e52e8b151c.png

\(\begin{array}{|rcll|} \hline AD^2 + 6^2 &=& 10^2 \\ AD^2 & = & 10^2 - 6^2 \\ AD^2 & = & 64 \\ \mathbf{AD} & \mathbf{=} & \mathbf{ 8 } \\\\ AB^2 &=& 15^2 + AD^2 \quad & | \quad AD^2 = 64 \\ AB^2 &=& 15^2 + 64 \\ AB^2 &=& 289 \\ \mathbf{AB} & \mathbf{=} & \mathbf{17 } \\\\ \sin(B) &=& \dfrac {AD} {AB} \\\\ \mathbf{\sin(B)} & \mathbf{=} & \mathbf{ \dfrac {8} {17}} \\ \hline \end{array}\)

3)
In triangle ABC, CA = 4sqrt2, CB = 4sqrt3, and angle A = 60 degrees. What is B in degrees?

sin-rule

\(\begin{array}{|rcll|} \hline \dfrac{\sin(B)} {4\sqrt{2}} &=& \dfrac{\sin(60^{\circ})} {4\sqrt{3}} \\\\ \dfrac{\sin(B)} { \sqrt{2}} &=& \dfrac{\sin(60^{\circ})} { \sqrt{3}} \quad & | \quad \sin(60^{\circ}) = \frac{ \sqrt{3} } {2} \\\\ \dfrac{\sin(B)} {\sqrt{2}} &=& \dfrac{ \sqrt{3} } {2\sqrt{3}} \\\\ \dfrac{\sin(B)} {\sqrt{2}} &=& \dfrac{ 1 } {2 } \\\\ \sin(B) &=& \dfrac{ \sqrt{2} } {2 } \\\\ \mathbf{B} & \mathbf{=} & \mathbf{45^{\circ}} \\ \hline \end{array}\)

There are 26 black cards and two red Kings, making 28 cards in total.  The probability of choosing one of these from all 52 cards is 28/52 or 7/13 ≈ 0.538

2)

Find AC. 

Picture: https://latex.artofproblemsolving.com/6/8/d/68dd918aac5d85a0b2a5b3b9c6a79eabd3669f01.png

sin-rule

\(\begin{array}{|rcll|} \hline \dfrac{AC}{\sin(45^{\circ})} &=& \dfrac{30} { \sin\Big(180^{\circ}-(75^{\circ}+45^{\circ}) \Big) } \quad & | \quad \small{\sin\Big(180^{\circ}-(75^{\circ}+45^{\circ}) \Big) = \sin(75^{\circ}+45^{\circ})} \\\\ \dfrac{AC}{\sin(45^{\circ})} &=& \dfrac{30} { \sin(75^{\circ}+45^{\circ}) } \\\\ \dfrac{AC}{\sin(45^{\circ})} &=& \dfrac{30} { \sin(120^{\circ}) } \\\\ AC &=& \dfrac{30\cdot \sin(45^{\circ})} { \sin(120^{\circ}) } \quad & | \quad \sin(45^{\circ}) = \frac{\sqrt{2}} {2} \\\\ AC &=& \dfrac{30\cdot \frac{\sqrt{2}} {2}} { \sin(120^{\circ}) } \quad & | \quad \sin(120^{\circ} = \frac{\sqrt{3}} {2} \\\\ AC &=& \dfrac{30\cdot \frac{\sqrt{2}} {2}} { \frac{\sqrt{3}} {2} } \\\\ AC &=& \dfrac{30\cdot \sqrt{2} } { \sqrt{3} } \cdot \frac{ \sqrt{3} } {\sqrt{3}} \\\\ AC &=& \dfrac{30\cdot \sqrt{2} \sqrt{3} } { 3 } \\\\ AC &=& 10\cdot \sqrt{6} \\ AC &=& 24.4948974278 \\ \hline \end{array}\)

1)

Find BC.

Picture: https://latex.artofproblemsolving.com/f/e/4/fe45662f573f87e30f9f30d0b1ef1f94b4c22fc2.png

cos-rule:

\(\begin{array}{|rcll|} \hline BC^2 &=& 6^2 +(7\sqrt{2})^2-2\cdot 6 \cdot7\sqrt{2}\cdot \cos(45^{\circ}) \quad & | \quad \cos(45^{\circ}) = \frac{\sqrt{2}}{2} \\ BC^2 &=& 36 + 49\cdot 2 -2\cdot 6 \cdot7\sqrt{2}\cdot \frac{\sqrt{2}}{2} \\ BC^2 &=& 36 + 98 -2\cdot 6 \cdot7 \cdot \frac{2}{2} \\ BC^2 &=& 36 + 98 -2\cdot 6 \cdot7 \\ BC^2 &=& 50 \\ BC &=&\sqrt{ 50} \\ BC &=&\sqrt{ 5^2\cdot 2} \\ BC &=& 5\sqrt{2} \\ \hline \end{array}\)

A function has a horizontal asymptote of y=-4 a vertical asymptote of x=3 and an -intercept at (1|0).

3²ⁿ + 3ⁿ - 2 = 0     n=?

You’re welcome. It’s great to have a student who actually participates in the Q&A. You may want to consider joining the forum. There are mathematicians here with advanced skills in most branches of mathematics. Some also have skills in writing computerized algorithms for “brute forcing,” sorting, and Monte Carlo simulations. 

GA

Mr. BB, the question asks for probability, not odds.

Your equation

nCr =20C3 =20! / {[20 - 3]!. 3!} =1,140 to 1

is SLOP! 

Why would you expect any student to understand this unconventional hodgepodge of SLOP?

In this case, nCr stands for “no Crap required.” None is required, because you’ve maxed it out!

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You can case work.

Case 1: WRW=4/9*5/8*3/7=15/126

Case 2: RWR=5/9*4/8*4/7=10/63

Total is 25/126.

(f•g)(x) --> f(g(x))
Basically, where there is an x, replace it with g(x), in this case x²+2

f(x)=x³-4x+2

f(x²+2)=(x²+2)³-4(x²+2)+2
           =x⁶+6x⁴+12x²+8-4x²-8+2
           =x⁶+6x⁴+8x²+2

Find the slope of the line going through (-5,1) and (1,1) using the slope formula:
m=(y₂-y₁)/(x₂-x₁₎
m=(1-1)/(1--5)
m=0/6

m=0

Parallel lines have the same slope, so the slope of the parallel line is 0.
Plug in (3,3) to y=mx+b as (x,y)
3=0(3)+b
b=3

line is y=3


 

Guest, I believe you misinterpreted the question. 

There are infinitely many ratios equivalent to the one given, \(\frac{4}{12}\) . Some examples are \(\frac{1}{3}\text{ and }\frac{2}{6}\text{ and }\frac{3}{9}\text{ and }\frac{5}{15}\) , just to name a few. 

Oh, I see it was just "intuitive" brute forcing :P. I guess I could've manually solved it that way but I was too lazy. I don't know what I was expecting, some ingenious algorithm of some sort, but I guess some things are simple enough to just calculate.

Thanks! 

All you need to add for your purpose are 4 numbers from the start and 4 numbers from the end:

7! + 8! + 9! + 10! +.................+ 2003! + 2004! + 2005! + 2006!=

214586420190..................40. All the other numbers in between are not going to change the 10th digit from the left, or "tens" digit from the right.

4/12 = 1/12 + 1/12 + 1/6 - This is one way.

But since 4/12 = 1/3, then you can have infinite number of solutions!

Example =1/3 =1/4 + 1/16 + 1/48...........and so on. 

Vertex form of a parabola is:
y=a(x-h)²+k; where (h,k) is the vertex

Vertex is given, plug them in.
y=a(x-10)²-7
 

Now we need to find the value of a.
Graph goes through (5,168), so plug those into x and y to solve for a.

168=a(5-10)²-7
175=a(-5)²

175=25a

a=7

With that, we now have the function of the parabola.
f(x)=7(x-10)²-7

∑[n!], n=7 to 2006 ≈2.145864201903033729347689996019583230087×10^5755

Counting from the left, the 10th digit is 1. But your question is: "What is the tens digit", which means the second last digit, which is a 4. Not sure which one you mean!.

98. There are a total of 59(<45) people, who are bald and not bald.

So, 18/59*100=30.5%(rounded)

make a 2-way table