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We have a right triangle, NSS....we know an angle and an opposite side to the angle

And we are looking for the side adjacent to the angle, x

[The adjacent side is the side next to the angle.....but not the hypotenuse....[ the side opposite the right angle]

The tangent of an angle = ratio of the its opposite side to its adjacent side....so

tan (17°) =    200 / x     multiply both sides by x

x * tan (17”)  = 200      divide bothsides by tan (17°)

x =   200 / tan (17°)  = 654 ft

Hope that helps, NSS....

3, 3, 3, 3, 3, 3, 3, 3, 2

We take this this in reverse order to find the continued fraction....it is actually not as hard as it seems...just a  (rather tedious) iterative process

So....We work from right to left

Take the last 2, invert it ⇒ 1/2....add it to the next-to-the-last 3   = 3 + 1/2 = 7/2 

Invert this = 2/7 and add it to the next 3 to the left = 3 + 2/7 = 23/7

Invert this =  7/23 and add it to the next 3 to the left   = 3 + 7/23 = 76/23

Invert this and add it to the next  3 on the left = 23/76 + 3 = 251/76

Invert this and add it to the next 3 on the left = 76/251 + 3  = 829/251

Invert this and add it to the next 3 on the left = 251 / 829 + 3 = 2738 / 829

Invert this and add it to the next 3 on the left = 829 / 2738 + 3 = 9043/2738

Invert this and add it to the last 3 on the left = 2738/9043 + 3  = 29867 / 9043  =  a / b

Nice, Heureka....!!!!

Use this formula to calculate the Present Value:
PV=0; P=250; R=0.04; N=3;    PV=P*(((1 + R)^N - 1)*((1 + R)^-N)* R^-1);print"PV =$",PV, where PV=Present Value, P=Payment, R=Interest Rate, N=Number of Periods.
@ 4% PV =$ 693.77
@ 6% PV =$ 668.25

Then I have to ask, what it means - cryptocode?

I suggested that ABCD is not equal to any of permutations (ABDC, BACD and so on).

The problem states that the letters can be arranged in any order.

I take that to mean that (w,x,y,z) and (x,y,z,w) and all it's permutations are the same code.

HERE WE ARE ..

If X is from the first set and from the other set W, X, Y are chosen, then possible cryptocodes are:

XWXY, XWYX, XXWY, XXYW, XYXW, XYWX, WXXY, WXYX, WYXX, YWXX, YXWX, YXXW - altogether 12 possibilities

If X is from the first set and from the other set we don't chose X at all, then we have 24 possibilities, for example:

if we chose X from the first set and W, Y, Z from the other set, we can have the following cryptocodes:

XWYZ, XWZY, XYWZ, XYZW, XZWY, XZYW, WXYZ, WXZY, WYXZ, WYZX, WZXY, WZYX,

YWXZ, YWZX, YXWZ, YXZW, YZXW, YZWX, ZWXY, ZWYX, ZYWX, ZYXW, ZXWY, ZXYW 

We can repeat until all letters have been chosen.

Every time when the four chosen letters are different, we have 24 possibilities, and if there are some letters twice, then we have 12 possibilites.

Result: there are altogether 312 possibilties!

Isn't it?

Thank you CPhill! I joined this site 19 days ago and it's awesome!

Thanks Melody! Wow you have a lot of hearts there 

I took your equation to mean:
0.97 =( 850-a^2) / 750     cross multiply
727.50 =850 - a^2
a^2 =850 - 727.50
a^2 = 122.5
a = +or- sqrt(122.5)
a=+or- 7sqrt(5/2)

Thank you😊

If a,b,c are the sides of a triangle then how do I prove that

\(\large{\dfrac {ab+bc+ca}{a^2+b^2+c^2} \geq \dfrac12 }\) ?

\(\begin{array}{|lrcll|} \hline 1. & a & \leq & b+c \quad | \quad \cdot a \\ & a^2 & \leq & (b+c)a \\ & \mathbf{a^2} & \mathbf{ \leq } & \mathbf{ ab+ca } \qquad (1) \\\\ 2. & b & \leq & a+c \quad | \quad \cdot b \\ & b^2 & \leq & (a+c)b \\ & \mathbf{b^2} & \mathbf{ \leq } & \mathbf{ab+bc} \qquad (2) \\\\ 3. & c & \leq & a+b \quad | \quad \cdot c \\ & c^2 & \leq & (a+b)c \\ & \mathbf{c^2} & \mathbf{ \leq } & \mathbf{ca+bc} \qquad (3) \\ \\ \hline \\ (1)+(2)+(3): & a^2+b^2+c^2 & \leq & ab+ca+ab+bc+ ca+bc \\ & a^2+b^2+c^2 & \leq & 2\cdot (ab+bc+ca) \quad | \quad : (a^2+b^2+c^2) \\ & 1 & \leq & 2\cdot \dfrac{ab+bc+ca} {a^2+b^2+c^2} \quad | \quad :2 \\ & \dfrac12 & \leq & \dfrac{ab+bc+ca} {a^2+b^2+c^2} \\ &\mathbf{ \dfrac{ab+bc+ca} {a^2+b^2+c^2}} & \mathbf{ \geq } & \mathbf{ \dfrac12 } \\ \hline \end{array}\)

The question was:

A balloon is losing 2 cubic inches of air per hour.

At what rate is it's radius decreasing when the balloon is 3 inches in radius.

\(\left( \begin{array}{cc} \{1,2,3,4,5\} & \{1,2,3,4,6\} \\ \{1,2,3,4,5\} & \{1,2,4,5,6\} \\ \{1,2,3,4,5\} & \{2,3,4,5,6\} \\ \{1,2,3,4,6\} & \{1,2,3,4,5\} \\ \{1,2,3,4,6\} & \{1,2,4,5,6\} \\ \{1,2,3,5,6\} & \{1,2,4,5,6\} \\ \{1,2,4,5,6\} & \{1,2,3,4,5\} \\ \{1,2,4,5,6\} & \{1,2,3,4,6\} \\ \{1,2,4,5,6\} & \{1,2,3,5,6\} \\ \{1,2,4,5,6\} & \{1,2,4,5,6\} \\ \{1,2,4,5,6\} & \{1,3,4,5,6\} \\ \{1,2,4,5,6\} & \{2,3,4,5,6\} \\ \{1,3,4,5,6\} & \{1,2,4,5,6\} \\ \{2,3,4,5,6\} & \{1,2,3,4,5\} \\ \{2,3,4,5,6\} & \{1,2,4,5,6\} \\ \end{array} \right)\)

This isn't quite correct.

You've treated the groups of three as permutations but the rows, columns and diagonals as combinations.

You need to be consistent in how you treat them.

If we went w/permuations we have 8 x 3! =48 different permuations for the rows, columns, diagonals

Then the probability becomes p = (48)/(504) = 2/21

If we use combinations we have 8 rows, columns, diagonals but then we have

9C3 = 84 possible arrangements of 3 darts on the board.

This gets us again a probability of p = 8/84 = 2/21

Using only the paths and the indicated directions, how many different routes are there from A to J?

\(\text{There are $\mathbf{22}$ routes from $A$ to $J$:} \\ \begin{array}{rl} 1 & \text{path of length } 3 \\ 6 & \text{paths of length } 4 \\ 10 & \text{paths of length } 5 \\ 5 & \text{paths of length } 6 \\ \end{array}\)

I'm afraid this isn't correct.

Ignoring duplicates for a moment there are

\(n = \dbinom{4}{1}\dbinom{4}{3} = 16\) different codes that are generated as described

There is however 1 code that is duplicated

(w,x,y,z) can be obtained by either (x)+(w,y,z) or (z) + (w,x,y)

So we end up with a total of 15 distinct codes.

i'm still confused...

Thus, (C) fits the option!

Sin (angle) = opp / hyp

Cosine = adj/hyp

Tan 17 = opp/adj

Tan 17 = 200/adj

adj = 200/Tan 17

See the detailed answer here. Hopefully, you can understand it:

https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.388792.html

This is from MATHCOUNTS, credit to them ;)

You are very welcome!

:P

REMEMBER:
sqrt361 =  + or - 19