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Not enough information

log_x {81} = log_2 {16}.

\(log_x {81} = log_2 {16}\\ log_x {3^4} = log_2 {2^4}\\ 4log_x {3} =4 log_2 {2}\\ log_x {3}=1\\ x=3 \)

\(\frac{32(cos(7\pi/4)+i sin(7\pi/4))} {4\sqrt2(cos(5\pi/6)+i sin(5\pi/6))}\\~\\ = \frac{32(e^{(7\pi/4)i})} {4\sqrt2(e^{(5\pi/6)i})}\\~\\ = \frac{8(e^{(7\pi/4)i-(5\pi/6)i})} {\sqrt2}\\~\\ = 4\sqrt2(e^{(7\pi/4)i-(5\pi/6)i})\)

you can check and finish it.

What is AS?

Hello Guest!

\(AS:12=(8+12):35\\ 35\cdot AS=12\cdot (8+12)=240\\ \color{blue}AS=6.\overline{857142}\)

  !

How many times does he label a point with the number 25?

Hello Guest!

12 times does he label a point with the number 25.

  !

\(\)

4 + 2 + 3 + 8==17 sum of all ratios

18 / 2 ==9 - share of the green ratio.

9  x  3 ==27 blue marbles.

I need to divide them

thank u

There are 8 possible combinations.

system

\(x(y + z) = 39\\ y(x + z) = 60\\ z(x + y) = 49\)
What is \(\mathbf{x}\)?

\(\begin{array}{|lrcll|} \hline & x(y + z) &=& 39 \\ (1) & xy+xz &=& 39 \\\\ &y(z + x) &=& 60 \\ (2)&yz + yx &=& 60 \\\\ &z(x + y) &=& 49 \\ (3) & zx+zy &=& 49\\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)-(2)+(3): &xy+xz-yz-yx +zx+zy&=& 39 -60+49 \\ &2xz &=& 28 \\ (4) &\mathbf{xz} &=& \mathbf{14} \\\\ (1)+(2)-(3): &xy+xz+yz+yx -zx-zy&=& 39 +60-49 \\ &2xy &=& 50 \\ (5) &\mathbf{xy} &=& \mathbf{25} \\\\ -(1)+(2)+(3): &-xy-xz+yz+yx +zx+zy&=& -39 +60+49 \\ &2yz &=& 70 \\ (6) &\mathbf{yz} &=& \mathbf{35} \\\\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \frac{(6)}{(4)}: & \dfrac{yz}{xz}&=&\dfrac{35}{14} \\\\ & \dfrac{y}{x}&=&\dfrac{35}{14} \\\\ & y &=&\dfrac{5}{2}*x \\\\ (5) & xy &=& 25\\\\ & x*\dfrac{5}{2}*x&=& 25 \\\\ & x^2 &=& \dfrac{2*25}{5} \\ & x^2 &=& 10 \\ & \mathbf{x} &=& \mathbf{\sqrt{10} }\\ \hline \end{array} \)

nose xd

What is the sum of all positive integer solutions less than or equal to 20 to the congruence 13(3x-2) == 26 - x (mod 8)?

13(3x-2) == 26 - x (mod 8)?

26-x (mod8) == 2-x (mod8)

13(3x-2) mod8 == 5(3x-2) (mod8) == 15x-10 (mod8) == -x-2 (mod8) == -2-x (mod 8)

So it doesn't like like there are any values of x.

I designed a little excel table to check

What is the x-coordinate of the point on the x-axis?

Hello Guest!

\(m_{g(x)}=\frac{y_2-y_1}{x_2-x_1}=\frac{6-0}{0+2}\\ m_{g(x)}=3\\ m_{f(x)}=-\frac{1}{3}\\ P_{center}(-1,\ 3)\)

\(f(x)=0=-\frac{1}{3}(x+1)+3\\ \frac{1}{3}(x+1)=3\\ \color{blue}x=8\)

The x-coordinate of the point on the x-axis is 8.

  !

Do you mean 

\(\frac{2}{3}* 1\frac{1}{2}\;\;? \)

I will answer for your inventivness in presentation.

\(\frac{2}{3}* 1\frac{1}{2}\ =​​\frac{2}{3}* \frac{3}{2}\\ \)

the threes cancel out and the 2s cancel out leaving you with   

\(\frac{1}{1}=1\)

g(x) = x^2 - 11x + 30

g(f(x)) = x^4 - 8x^3 + 11x^2 + 20x + 6

Let  f(x)= x^2 +Bx+C

then

g(f(x))= (x^2 +Bx+C)^2 - 11(x^2 +Bx+C) +30

Now just expeand it and equate the coefficients for find B and C ,

It is not that difficult.

Can you check your input please.

You have x^2 twice in the second function.

85  -  88 == - 3 - this is the common difference

- 512 ==88  + (- 3 *( N - 1))

- 512 ==88  + - 3N + 3

- 512 - 88 - 3 ==- 3N

- 603 == - 3N

N ==- 603 / -3

N ==201 - this term number gives - 512

So: 201 - 1 ==200 terms before you get to - 512

Solve 2^(16x) = 16^(2x).

Hello Guest!

\(2^{16x} = 16^{2x}\\ 2^{16x} = 2^{4\cdot 2x}\\ 16x=8x\\ 8x=0\\ \color{blue}x=0 \)

  !

And welcome to the web2.0 forum.   I hope you have fun and learn lots.

Like this.

\(tan17=\frac{x}{y}\qquad (1)\\ tan 21.8 = \frac{x+8}{y}\qquad (2)\)

solve simultaneously.

answer is 25.94

but how

Thought it was 103...problem don't make sense xD

24! + 12!= 2^10 * 3^5 * 5^2 * 7 * 11 * 821 * 11087 * 142302163

Solve the following system:

{a = -a + 2 + sqrt(5) | (equation 1)

b = -b + 2 - sqrt(5) | (equation 2)

c = -c + 2 + sqrt(5) | (equation 3)

d = -d + 2 - sqrt(5) | (equation 4)

Express the system in standard form:

{2 a+0 b+0 c+0 d = 2 + sqrt(5) | (equation 1)

0 a+2 b+0 c+0 d = 2 - sqrt(5) | (equation 2)

0 a+0 b+2 c+0 d = 2 + sqrt(5) | (equation 3)

0 a+0 b+0 c+2 d = 2 - sqrt(5) | (equation 4)

Divide equation 4 by 2:

{2 a+0 b+0 c+0 d = 2 + sqrt(5) | (equation 1)

0 a+2 b+0 c+0 d = 2 - sqrt(5) | (equation 2)

0 a+0 b+2 c+0 d = 2 + sqrt(5) | (equation 3)

0 a+0 b+0 c+d = 1 - (sqrt(5))/(2) | (equation 4)

Divide equation 3 by 2:

{2 a+0 b+0 c+0 d = 2 + sqrt(5) | (equation 1)

0 a+2 b+0 c+0 d = 2 - sqrt(5) | (equation 2)

0 a+0 b+c+0 d = (sqrt(5) + 2)/(2) | (equation 3)

0 a+0 b+0 c+d = 1 - sqrt(5)/2 | (equation 4)

Divide equation 2 by 2:

{2 a+0 b+0 c+0 d = 2 + sqrt(5) | (equation 1)

0 a+b+0 c+0 d = 1 - (sqrt(5))/(2) | (equation 2)

0 a+0 b+c+0 d = 1/2 (2 + sqrt(5)) | (equation 3)

0 a+0 b+0 c+d = 1 - sqrt(5)/2 | (equation 4)

Divide equation 1 by 2:

{a+0 b+0 c+0 d = (sqrt(5) + 2)/(2) | (equation 1)

0 a+b+0 c+0 d = 1 - sqrt(5)/2 | (equation 2)

0 a+0 b+c+0 d = 1/2 (2 + sqrt(5)) | (equation 3)

0 a+0 b+0 c+d = 1 - sqrt(5)/2 | (equation 4)

Collect results:

a = 1/2 (2 + sqrt(5))

b = 1 - sqrt(5)/2

c = 1/2 (2 + sqrt(5))

d = 1 - sqrt(5)/2      If you multiply them together, you get:abcd =1/16