All Answers 352766 Answers

95 degrees  + 3/60 degrees + 55/3600 degrees = 95.0652778 degrees

95.0652778 degrees x  pi radians/180 degrees = 1.659 R = ~~ 1.66 Radians

Use Pythagorean Theorem:

square the shorter side lengths and add them together....then see if that sum equals the square of the longer side.....try it!

8 -15 - 17   IS a right triangle....you try the others

8^2 + 15^2  =? 17^2

64 + 225 =? 289

289 = 289    yes, it is a right triangle !

Cosine is adjacent/hypotenuse    you are given opposite and adjacent....

you can find the tan of the angle and then find the cosine of that angle...or calculate the hypotenuse with pythag theorem

hyp^2 = 75^2 +50^2     hyp = 90.139

then cos = adj/hyp = 75/90.138 = .83

OK.....glad to help   !!!!

Thank you so much for the help, I struggled where it ask to show the slopes are same. But after seeing your answer it made more sense. 

Thanks

I just added this to our list of useful calculators. 

5) The product of all digits of positive integer M is 105. How many such Ms are there with distinct digits?

Here's my best attempt.....

105 factors as  3 * 5 * 7

So...we have the following possibilities

357    

375

537

573

735

753

So...6 M's

One way to enter    6C5 is     nCr(6,5)

A = 500    given

find t

500 = 1562.5 * 10^(-.1t)

500/1562.5  = 10^(-.1t)

.32 = 10^(-.1t)                log both sides

log .32 = -.1t

log .32 / (-.1)  = t      (4.95 years)

-10 log (.32)

This problem becomes easier if we lay it out like this :

Let D  = (0,0)

A  =  (-24 cos60, 24 sin 60)  =  (-12, 12√3)

C = (24 cos 60, 24 sin 60)  =  (12, 12√3)

Let B = (0, y)

To  find y....we can use the square of the distance formula

(12 - 0)^2  + (y - 12√3)^2  = 13^2

12^2 + (y - 12√3)^2  =  13^2

(y - 12√3)^2  = 13^2 - 12^2

(y - 12√3)^2  = 25       take the square root of both sides

y - 12√3  = 5

y =  = 5 + 12√3

So....B  = (0, 5 + 12√3)

X =  [  (12 + 0)/2,  (12√3 + 5 + 12√3)/2 ]  =   (6, 2.5 +12√3)

Y  =  (-6, 6√3)

So XY^2   =    (6 - - 6)^2  + ( 2.5 + 12√3  - 6√3)^2  =

(12)^2  + (2.5 + 6√3)^2  =

144 + 6.25 + 30√3 + 108  =

258.25 + 30√3  units  ≈ 310.212  units

x = 13.7 * sin 33.2  / sin 45.6 = 10.49951  ~~ 10.50  (rounded)

dramas 

2/3

3+4+5 = 12 movies total       3+ 5 = 8 movies are NOT comedies      8/12 = 2/3

Yes. 

If you want an old question temporarily opened you can privately message a moderator and request it. 

In convex quadrilateral abcd,ab=bc=13 , cd=da=24, and angle D= 60 degrees. 

Points X andY  are the midpoints of BC and  DA respectively. Compute XY^2 (the square of the length of  XY).

\(\text{Let $\angle DAC =\angle ACD = 60^\circ $} \\ \text{Let $\angle ACY = \dfrac{\angle ACD}{2} = 30^\circ $} \\ \text{Let $ AD=AC=CD =24 $} \\ \text{Let $ CY = u $} \)

\(\mathbf{u=\ ?}\)

\(\begin{array}{|rcll|} \hline u^2+12^2 &=& 24^2 \\ u^2 &=& 24^2-12^2 \\ \mathbf{u^2} &\mathbf{=}& \mathbf{432} \\ \hline \end{array}\)

cos-Rule \(\mathbf{\cos(B)=\ ?}\):

\(\begin{array}{|rcll|} \hline 24^2 &=& 13^2+13^2-2\cdot 13 \cdot 13 \cdot \cos(B) \\ \ldots \\ \mathbf{\cos(B)} &\mathbf{=}& \mathbf{1-\dfrac{24^2}{2\cdot 13^2}} \\ \hline \end{array}\)

\(\mathbf{\cos\left(\dfrac{B}{2}\right)=\ ?}: \)

\(\begin{array}{|rcll|} \hline \cos(B) &=& 2\cos^2\left(\dfrac{B}{2}\right) - 1 \\ 2\cos^2\left(\dfrac{B}{2}\right) &=& 1+ \cos(B) \quad | \quad \mathbf{\cos(B) = 1-\dfrac{24^2}{2\cdot 13^2} } \\ 2\cos^2\left(\dfrac{B}{2}\right) &=& 1+ 1-\dfrac{24^2}{2\cdot 13^2} \\ 2\cos^2\left(\dfrac{B}{2}\right) &=& 2-\dfrac{24^2}{2\cdot 13^2} \\ \cos^2\left(\dfrac{B}{2}\right) &=& 1-\dfrac{24^2}{2^2\cdot 13^2} \\ \cos^2\left(\dfrac{B}{2}\right) &=& \dfrac{2^2\cdot 13^2-24^2}{2^2\cdot 13^2} \\ \cos^2\left(\dfrac{B}{2}\right) &=& \dfrac{10^2}{2^2\cdot 13^2} \\ \cos\left(\dfrac{B}{2}\right) &=& \dfrac{10}{2\cdot 13} \\ \mathbf{\cos\left(\dfrac{B}{2}\right)} &\mathbf{=}& \mathbf{\dfrac{5}{13}} \\ \hline \end{array}\)

\(\mathbf{\sin(\alpha)=\ ?}\):

\(\begin{array}{|rcll|} \hline 180^\circ &=& B + 2\alpha \\ \ldots \\ \alpha &=& 90^\circ -\dfrac{B}{2} \\ \sin(\alpha) &=& \sin\left( 90^\circ -\dfrac{B}{2} \right) \\ \sin(\alpha) &=& \cos\left(\dfrac{B}{2} \right) \\ \mathbf{\sin(\alpha)} &\mathbf{=}& \mathbf{\dfrac{5}{13}} \\\\ \cos(\alpha) &=& \sqrt{1-\sin^2(\alpha)} \\ &=& \sqrt{1-\dfrac{5^2}{13^2}} \\ &=& \sqrt{ \dfrac{13^2-5^2}{13^2}} \\ &=& \sqrt{ \dfrac{12^2}{13^2}} \\ \mathbf{\cos(\alpha)} &\mathbf{=}& \mathbf{\dfrac{12}{13}} \\ \hline \end{array}\)

cos-Rule \(\mathbf{\overline{XY}^2=\ ?}\):

\(\begin{array}{|rcll|} \hline \overline{XY}^2 &=& 6.5^2+u^2-2\cdot 6.5 \cdot u \cdot \cos(\alpha+30^\circ) \\ &=& 6.5^2+432-13 \cdot u \cdot \cos(\alpha+30^\circ) \\ &=& 474.25-13 \cdot \sqrt{432} \cdot \cos(\alpha+30^\circ) \\ &=& 474.25-13 \cdot \sqrt{432} \cdot \Big( \cos(\alpha) \cos(30^\circ)-\sin(\alpha)\sin(30^\circ ) \Big) \\ &=& 474.25-13 \cdot \sqrt{432} \cdot \Big( \dfrac{12}{13}\cdot \dfrac{\sqrt{3}} {2} -\dfrac{5}{13}\cdot \dfrac{1} {2} \Big) \\ &=& 474.25-13 \cdot \sqrt{432} \cdot \Big( \dfrac {12 \sqrt{3}-5}{2\cdot 13} \Big) \\ &=& 474.25- \sqrt{432} \cdot \Big( \dfrac {12 \sqrt{3}-5}{2 } \Big) \\ &=& 474.25- \sqrt{\dfrac{432}{4}} \cdot \left( 12 \sqrt{3}-5 \right) \\ &=& 474.25- \sqrt{108} \left( 12 \sqrt{3}-5 \right) \\ &=& 474.25- \sqrt{4\cdot 27} \left( 12 \sqrt{3}-5 \right) \\ &=& 474.25- \sqrt{2^2\cdot 3^2\cdot 3} \left( 12 \sqrt{3}-5 \right) \\ &=& 474.25- 6\sqrt{ 3} \left( 12 \sqrt{3}-5 \right) \\ &=& 474.25- 6\sqrt{ 3} \cdot 12 \sqrt{3} -5\cdot 6\sqrt{ 3} \\ &=& 474.25- 72\ \cdot 3 +30\sqrt{ 3} \\ &=& 474.25- 216 +30\sqrt{ 3} \\ &=& 258.25 + 30\sqrt{ 3} \\ &=& 258.25 + 51.9615242271 \\ \mathbf{\overline{XY}^2} &\mathbf{=} & \mathbf{310.211524227} \\ \hline \end{array}\)

It seems to me that this is going to be the same as rolling a dice 5 times and finding the probability that 4 is rolled any 2 times in a row.

call a 4 a success and any other a failure then   for any individual roll   P(S)=1/6     P(F)=5/6     

44444  Any 5 fours        1 way                                                   (1/6)^5

4444  any 4 fours           5 ways                                                 5*(1/6)^4*(5/6)

444      an 3 fours          5C3 ways = 10 ways                           10*(1/6)^3*(5/6)^2

except   4*4*4     no but all other 3 fours ok.    -1 way                (1/6)^3*(5/6)^2

No fours 1 way    (5/6)^5                                                               

P(4*4**)=P(4**4*)=P(4***4)=P(*4*4*)=P(*4**4)=P(**4*4) = (1/6)^2*(5/6)^3

\(\text {P(rolling 2 fours together in 5 rolls ) }\\ =\text{1-P(no fours)-P(1 four)-P(2 fours but not together) - P(3 fours but none together) } \\=1-(\frac{5}{6})^5- 5(\frac{1}{6})^1(\frac{5}{6})^4 - 6(\frac{1}{6})^2(\frac{5}{6})^3 - (\frac{1}{6})^3(\frac{5}{6})^2\)

1-(5/6)^5-5*(1/6)^1*(5/6)^4-6*(1/6)^2*(5/6)^3-(1/6)^3*(5/6)^2 = 0.0965792181069959   \(=\frac{751}{7776}\)

2 fours in the right places

44***

*44**

**44*

***44

P(2 fours in the right places) = 4*(1/6)^2(5/6)^3

OR

P(rolling 2 fours together)

=P(5 fours)+P(4 fours)+ P(3 fours) - P(4*4*4) +P(2 fours in the right places)

(1/6)^5+5*(1/6)^4*(5/6)^1+10*(1/6)^3*(5/6)^2-(1/6)^3*(5/6)^2+4*(1/6)^2(5/6)^3 = 0.0965792181069959    \(=\frac{751}{7776}\)

That is excellent.

I did it 2 different ways and got the same answer both times :)

To two decimal places, the answer is 10.50.

You both need to check your arithmetic.

Suppose that ABCD is a trapezoid in which AD || BC. Given AC perpendicular CD, AC bisects angle BAD, and the area of ABCD is 42, then compute the area of triangle ACD.

On the xy-plane, the origin is labeled with an M.

The points (1,0), (-1,0), (0,1), and (0,-1) are labeled with A's.

The points (2,0), (1,1), (0,2), (-1, 1), (-2, 0), (-1, -1), (0, -2), and (1, -1) are labeled with T's.

The points (3,0), (2,1), (1,2), (0, 3), (-1, 2), (-2, 1), (-3, 0), (-2,-1), (-1,-2), (0, -3), (1, -2), and (2, -1) are labeled with H's.

If you are only allowed to move up, down, left, and right, starting from the origin,

how many distinct paths can be followed to spell the word MATH?

\(\begin{array}{|lclcl|} \hline \text{Let } d &-& \text{ down } &=& 0 \\ \text{Let } r &-& \text{ right } &=& 1 \\ \text{Let } u &-& \text{ up } &=& 2 \\ \text{Let } l &-& \text{ left } &=& 3 \\ \hline \end{array}\)

From M to "down-A", all distinct paths

are all three digit numbers to base 4:

\(\begin{array}{|r|r|l|} \hline base_4 &\text{path} & \text{spell the word MATH} \\ \hline 000 & ddd & \checkmark \\ 001 & ddr & \checkmark \\ 002 & ddu & \\ 003 & ddl & \checkmark \\ 010 & drd & \checkmark \\ 011 & drr & \checkmark \\ 012 & dru & \\ 013 & drl & \\ 020 & dud & \\ 021 & dur & \\ 022 & duu & \\ 023 & dul & \\ 030 & dld & \checkmark \\ 031 & dlr & \\ 032 & dlu & \\ 033 & dll & \checkmark \\ \hline && 7 \text{ distinct paths to spell the word MATH } \\ \hline \end{array} \)

For reasons of symmetry:

\(\begin{array}{|lcll|} \hline \text{From M to "down-A" there are } 7 \text{ distinct paths to spell the word MATH } & base_4 \quad 000 \to 033 \\ \text{From M to "right-A" there are } 7 \text{ distinct paths to spell the word MATH }& base_4 \quad 100 \to 133 \\ \text{From M to "up-A" there are 7 } \text{ distinct paths to spell the word MATH } & base_4 \quad 200 \to 233 \\ \text{From M to "left-A" there are } 7 \text{ distinct paths to spell the word MATH } & base_4 \quad 300 \to 333 \\ \hline \end{array} \)

\(4\times 7 = 28\) distinct paths can be followed to spell the word MATH

\(y=(\sqrt2)^{(\frac{7\pi i}{4})}\\ y=2^{(\frac{7\pi }{8}i)}\\ log y=log 2^{(\frac{7\pi }{8}i)}\\ ln y=(\frac{7\pi }{8}i) ln 2\\ e^{ln y}=e^{(\frac{7ln 2\pi }{8}i) }\\ y=e^{(\frac{7(ln 2)\pi }{8})i }\\ y=cos(\frac{7(ln 2)\pi }{8})+isin(\frac{7(ln 2)\pi }{8})\)

You had better check that very carefully.

there are 3 R's, 2 E's, 2 A's

some of the permutations will be identical because of this

The forum locks questions to prevent spam after they are posted for a certain period of time. This keeps users and the site itself safe and managable. Furthermore, people can use old post to farm experience, as was a problem with Reddit a while back, I remember when the site was down :)