Проблемы по математике за 2-ой класс.

[bigfatred]

1. Convert the following duodecimal number to hex: 0.BA

0.BA(base-12) ---->into base-10:

.71/72= 0.986111111

0.986111111(base 10)----->into base 16:

0. FC7

В учебнике дан ответ 0.FC8. После семи тысячной идет единица. Не знаю как восьмерка возможна.

[Буржуй]

0;BA in duodecimal is for sure 0.FC71C71C71C7 in hex but it aint 2nd grade math :rofl:

[bigfatred]

0;BA in duodecimal is for sure 0.FC71C71C71C7 in hex but it aint 2nd grade math :rofl:

thank you. "problem" destroyed.

[bigfatred]

3/8 of A with B and C in the ratio 1:2

Find A:B:C

(1st grade math)

[bigfatred]

Мне кажется график части одной функций выглядит как-то не так. Та часть дана таким образом.


x^2+x+1: x<1, x=1 for [-3, 4] with intervals of 0.5

x,y pairs (-3, 7), (-2, 3), (-1,1)

Do these ordered pairs look wrong to you? If you look at this part of the graph in my textbook it goes through different points.

[bigfatred]

Please, take a look at this one below:

log(base A) of B= 1/[log(base B) of A]

because: if log(base B) of A=C, then B^C=A and so B=A^1/C

Hence, log(base A) of B= 1/C= 1/[log(base B) of A]

I don't see how introducing C as equal to a different base and then carrying out a bunch of algebraic manipulations prove anything. Am I missing something here? Thank You

[Quanty]

пусть ребенок сам домашнюю работу делает.

[Quanty]

they are just using definition of a log. ...they don't introduce C "as equal to a different base"...

[Буржуй]

пусть ребенок сам домашнюю работу делает.

это не задачки первого второго класа

[bigfatred]

they are just using definition of a log. ...they don't introduce C "as equal to a different base"...

Yeah I just realized that.

If A=B^C, then C=log(base B)A

Solved.