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3 years ago

8

List the first twenty counting numbers in the indicated base below. twelve (Only digits 0, 1, 2, 9, A, B are used in base twelve

.) What are the first twenty counting numbers in base twelve? (Use a comma to separate answers as needed.)

1 answer:

kicyunya [14]3 years ago

6 0

Answer:

0, 1 , 2, 3, 4, 5, 6, 7, 8, 9, A, B, 10, 11, 12, 13, 14, 15, 16, 17, 18

Step-By-Step Explanation:

i will go from 0 to 20:

0: 0

1: 1

2: 2

3: 3

4: 4

5: 5

6: 6

7: 7

8: 8

9: 9

10: A (10 is another 'digit')

11: B

12: 10 (12 = 1*12^1 + 0* 12^0)

13: 11 (13 = 1*12^1 + 1*12^0)

14: 12 (14 = 1*12^1 + 2*12^0)

15: 13

16: 14

17: 15

18: 16

19: 17

20: 18

Just remember to use the notation A and B after the digit nine, for example

22: 1A

23: 1B

24: 20

In other words, the answer is 0, 1 , 2, 3, 4, 5, 6, 7, 8, 9, A, B, 10, 11, 12, 13, 14, 15, 16, 17, 18

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Step-by-step explanation:

Substitute the equation and values into the average rate of change formula.

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3 years ago

The line AB has midpoint (2,5).<br> A has coordinates (1, 2).<br> Find the coordinates of B.

Gekata [30.6K]

Answer:

$X_m = \frac{A_x +B_x}{2}= \frac{1+B_x}{2}= 2$

And we can solve for $B_x$ and we got:

$1+B_x = 4$

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And we can solve for $B_x$ and we got:

$2+B_y = 10$

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Step-by-step explanation:

For this case we know that the midpoint for the segment AB is (2,5)

And we know that the coordinates of A are (1,2)

We know that for a given segment the formulas in order to find the midpoint are given by:

$X_m = \frac{A_x +B_x}{2}= \frac{1+B_x}{2}= 2$

And we can solve for $B_x$ and we got:

$1+B_x = 4$

$B_x = 3$

$Y_m = \frac{A_y +B_y}{2}= \frac{2+B_y}{2}= 5$

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3 years ago

What is a solution to the equation 3 / m + 3 - M / 3 - M equals m^2 + 9 / m^2-9?

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Answer: Last option.

Step-by-step explanation:

Given the equation:

$\frac{3}{m+3}-\frac{m}{3-m}=\frac{m^2+9}{m^2-9}$

Follow these steps to solve it:

- Subtract the fractions on the left side of the equation:

$\frac{3(3-m)-m(m+3)}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{9-3m-m^2-3m}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}$

- Using the Difference of squares formula ( $a^2-b^2=(a+b)(a-b)$ ) we can simplify the denominator of the right side of the equation:

$\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{(m+3)(m-3)}$

- Multiply both sides of the equation by $(m+3)(3-m)$ and simplify:

$\frac{(-m^2-6m+9)(m+3)(3-m)}{(m+3)(3-m)}=\frac{(m^2+9)(m+3)(3-m)}{(m+3)(m-3)}\\\\-m^2-6m+9=\frac{(m^2+9)(3-m)}{(m-3)}$

- Multiply both sides by $m-3$ :

$(-m^2-6m+9)(m-3)=\frac{(m^2+9)(3-m)(m-3)}{(m-3)}\\\\(-m^2-6m+9)(m-3)=(m^2+9)(3-m)$

- Apply Distributive property and simplify:

$(-m^2-6m+9)(m-3)=(m^2+9)(3-m)\\\\-m^3-6m^2+9m+3m^2+18m-27=3m^2+27-m^3-9m\\\\-m^3-3m^2+27m-27+m^3-3m^2+9m-27=0\\\\-6m^2+36m-54=0$

- Divide both sides of the equation by -6:

$\frac{-6m^2+36m-54}{-6}=\frac{0}{-6}\\\\m^2-6m+9=0$

- Factor the equation and solve for "m":

$(m-3)^2=0\\\\m=3$

In order to verify it, you must substitute $m=3$ into the equation and solve it:

$\frac{3}{3+3}-\frac{3}{3-3}=\frac{3^2+9}{3^2-9}\\\\\frac{3}{6}-\frac{3}{0}=\frac{18}{0}$

<em>NO SOLUTION</em>

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3 years ago

Solve -5y+6=-9 pls help me with this

alexandr402 [8]

Answer:

-5y + 6 = -9

y = 3

Step-by-step explanation:

subtract 6 from both sides

-5y + 6 - 6 = -9 - 6

simplify

-5y = -15

divide both sides by -5

(-5y)/5 = (-15)/5

simplify

y = 3

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3 years ago