Since you mentioned the fundamental counting principle, I assume you are looking for the total number of possibilities.

If you are picking a month, there are 12 options.
If you are selecting a day of the week, there are 7 options.

If you are doing both, you multiply the two numbers together.
12 x 7 = 84

8 0

3 years ago

Solve the equation for y.<br><br> 8x – 9y = 11

SVETLANKA909090 [29]

8x-9y=11

Add 9y to both sides

8x = 11 + 9y

Subtract 11 on both sides

8x - 11 = 9y

Divide 9 on both sides

$\boxed{y = \frac{8x-11}{9} }$

5 0

3 years ago

Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}

miskamm [114]

Answer:

$-3+i\sqrt{3} , 1+\sqrt{3}$

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

$\alpha = x+iy\\\beta = x-iy$

since they are conjugates

$\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}$

$\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}$

Imaginary part of the above =0

i.e. $\sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1$

So the value of alpha = $-3+i\sqrt{3} , 1+\sqrt{3}$

3 0

3 years ago