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cos (x)  = .35

arccos (.35)  = x ≈   69.5°

Answer: 548 for the first one, 720 for the second one.

Solution:

Case 1: The first digit is a prime number; none of the rest are.

4 (there are four primes 1-9) x 6 (6 non-primes) x 5 (5 non-prime and non-whatever-the-second-digit-was numbers) = 120 

Case 2: The second digit is a prime number; none of the rest are.

5 x 4 x 5 = 100

Case 3: The third digit is a prime number; none of the rest are.

5 x 5 x 4 = 100

Case 4: The first two are prime; the third isn't.

4 x 3 x 6 = 72

Case 5: The last two are prime; the first isn't.

5 x 4 x 3 = 60

Case 6: The middle digit isn't prime; the rest are.

4 x 6 x 3 = 72

Case 7: All three digits are prime.

4 x 3 x 2 = 24

Adding these all up gives a sum of 548.

Tweaking the formulas used a little bit to answer the second question gives an answer of 720.

There's probably a more efficient way than this, but it does beat listing them out (if it is right, that is). 

sorry, it is incorrect, im not sure why tho

Equation 1:   \(2x+y=-7\)

Rewrite Equation 2 as follows: \(4x=4+2y \iff 2x=2+y\)

Substitute Equation 2 in 1:

\(2+y+y=-7 \implies y=-\dfrac{9}{2}\)

Thus, \(2x=2-\dfrac{9}{2}=-\frac{5}{2} \implies x=-\dfrac{5}{4}\)

I hope this helps!

Thanks Doremy.    

I was confused because I had not read the question properly.

Wait... I made an error... 

The probability is \((5^{394} \times 2^{96}) \times 2^{-392} \times 3^{-470} = 5^{384} \times 2^{-296} \times 3^{-470} = \color{brown}\boxed{-382}\)

Builderboi has a great way to solve it but sometimes you don't have much of those formulas memorized.

Use polynomial construction.

a + b + c = 1

ab + ac + bc = -1

abc = -1

P(x) = x^3 - x^2 - x + 1

a, b, c are the roots of this polynomial. 

a^3 = a^2 + a - 1

b^3 = b^2 + b - 1

c^3 = c^2 + c - 1

a^3 + b^3 + c^3 = (a^2 + b^2 + c^2) + (a + b + c) - 3

(a + b + c)^2 - 2(ab + ac + bc) = a^2 + b^2 +  c^2 = 3

a^3 + b^3 + c^3 = 1.

We can use the fact that for any complex number a, (a)(cong(a)) = |a|^2.

\(|5z + w|^2 = (5z + w)(5\overline{z} + \overline{w}) = 121 \)

\(25|z|^2 + |w|^2 + 5(\overline{z}w + \overline{w}z) = 121\)

\(5(\overline{z}w + \overline{w}z) = 47\)

\(|3z + 4w|^2 = (3z + 4w)(3\overline{z} + 4\overline{w}) \)

\(=793 + 12(\overline{w}z + \overline{z}w)\)

\(=793 + 564/5 \)

Take the square root of that value and you will the answer.

OK

We have that

-6t^2 + 51t    =   108         divide through by -3

2t^2  - 17t  =  -36

2t^2 - 17t + 36 =  0      see if we  can  factor   (you could use the Q Formula but I like to  factor, if possible )

(2t - 9) ( t - 4)  =  0

2t - 9  = 0                                 or            t  - 4   =  0

2t = 9                                                       t = 4 sec

t = 9/2 sec = 4.5 sec

It first reaches 108 ft  at 4 sec

-6t^2 + 51t

o holld up

Welcome....!!!

BTW -  no equation shown for the second one

thxs :D

(x - 2) ( 2x + 5)  = 8x - 6

2x^2 - 4x + 5x - 10 = 8x -6

2x^2 + x -10 =  8x - 6       rearrange as

2x^2 - 7x - 4  = 0

(2x + 1) ( x - 4)  =  0

The second factor set = 0  is what we want 

x - 4 =  0 

x  = 4

Each side of the cube is \(\sqrt[3]{1000} = 10\).

There are 512 cubes inside that have no faces painted, so the probability is \(1^{512} = 1 \)

There are 8 corners, each with a probability of \({3 \over 6} = {1 \over 2} = 2^{-1}\), so the probability is \({1 \over 2}^8 = {1 \over 2^8}\)

There are \(8 \times 12 = 96\) edge pieces (there are 12 edges in a cube, each edge has 8), and each cube has a probability of \({4 \over 6} = {2 \over 3}\), so the probability is \({2 \over 3}^{96} = {2^{96} \over 3^{96}}\)

There are \(8 \times 8 \times 6 = 384\) center cubes, each with a probability of \({5 \over 6} \), so the probability is \(\large{{5 \over 6}^{384} = {5^{384} \over 6^{384}} = {{5^{384}} \over 2^{384} \times 3^{384}}}\). 

So, the probability is \(\large{{1 \over 2^8} \times {2^{96} \over 3^{96}} \times {5^{384} \over 2^{384} \times 3^{384}} = {5^{384} \times 2^{96} \over 2^{392} \times 3^{470}} = {(5^{384} \times 2^{96})} \times 2^{-392} \times 3^{470} = 5^{384} \times 2^{-296} \times 3^{470}} \)

Thus, \(a + b + c = \color{brown}\boxed{558}\)

3ax + 3y   =  5b

2x + y = 5   multiply through by 3  →  6x + 3y  = 15

To have infinite solutions the equations must be the same ......this implies that

3a =  6         and           5b  = 15  

a  =   2                              b =  3

x^2 + y^2 + z^2 + 2(xy + yz + zx) = (x + y + z) ^2

= 21 = (x + y + z)^2

x + y + z = sqrt(21)

(Make sure you know why it's positive sqrt(21)!)

No graph shown  ....

Rearrange as

sqrt (10x -17)  = sqrt (5x - 102)     square both sides

10x  -  17 =   5x  -102

10x  -5x  =  -102 + 17

5x  =  -85

x = -85/ 5  =  -17

Note that

10 (-17) - 17 = - 187

And

5(-17) - 102 =  -187

So  x = -17  is the  solution (assuming that complex forms  are OK )  and we actually have

sqrt ( -187) - sqrt (-187)  = 0

sqrt (187) i  -  sqrt (187) i  =  0

 i ( sqrt (187) - sqrt (187) )  = 0

i (0)  = 0   

0i = 0   

0 = 0   is true

AG = sqrt (18^2 - 12^2)  =  sqrt (324 - 144)  = sqrt (180)  = 6sqrt (5)

Call the intersection of GF and DC = H

Let  DG  = x      DH  = y

Triangles DGH  and ABG   are  similar 

So

DG/ DH   = AB / AG

x / y  =  12 / 6 sqrt 5

x / y =   2 /sqrt 5

y =  [ sqrt (5) x / 2 ]  =  DH

And triangles FCH and DGH and  ABG are similar

CH = 12 - y =    12 - [sqrt(5) x / 2]  

FH / CH  = AG / BG

FH / CH  = 6sqrt (5) / 18

FH / CH  = sqrt (5) / 3

FH = CH * sqrt (5) /3

FH  =  (12 - sqrt (5)x / 2 ) (sqrt (5) /3)   =  (24sqrt (5)  - 5x) / 6 

DG^2 + DH^2  =  FH^2

x^2  + (sqrt(5)x /2)^2  =  [   (24sqrt (5)  - 5x) / 6 ]^2

Solving for positive  x produces  x =  (12/7)sqrt (5)   = DG

AD = 

 DG + AG  = 

(12/7)sqrt (5)  + 6sqrt (5) =   

(12/7 + 6)sqrt (5) =

(54/7) sqrt (5)

a*(d+3+f+g)

b(d+e+f+g)

c(d+e+f+g)

12

wrong question

Q

          15

P                R

tan R =  5 cos Q                          

QP / PR  = 5  (QP / QR)

QP / PR  = 5QP / 15        

1/ PR =  5/ 15

1/ PR  = 1/3

PR  = 3

So

PQ = sqrt [  QR^2  - PR^2  ] =   sqrt [ 15^2  - 3^2]  = sqrt [ 225 - 9 ] = sqrt [216] = sqrt [36 * 6 ] = 6 sqrt (6)

4x  - 6y    = -2    mult through by 3    →   12x - 18y  =  -6

kx  - 9y    =  3    mult through by 2    →   2kx  - 18y =    6

For this to have no real solutions 

2kx   =   12x

2k = 12

k = 6