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AleksandrR [38]
3 years ago
5

How do I mutiply 14(a+5+6b)

Mathematics
2 answers:
OLEGan [10]3 years ago
6 0

Answer:

14a + 70 + 84b

Step-by-step explanation:

Each term in the parenthesis is multiplied by 14, that is

= (14 × a) + (14 × 5) + (14 × 6b)

= 14a + 70 + 84b

padilas [110]3 years ago
4 0
Multiply 14 to each term: 14a+70+84b
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Use the points (0, 4) and (-2, 3) to find the slope of the line and give the equation of this line.​
Sauron [17]

Answer:

Choice C:   m=\frac{1}{2}; y=\frac{1}{2} x+4

Step-by-step explanation:

To find your slope do:

\frac{y^2-y^1}{x^2-x^1}

\frac{3-4}{-2-0}=\frac{-1}{-2} =\frac{1}{2}

that's how you get 1/2

4 is just the y-int.

6 0
2 years ago
How do i write 3% as a fraction​
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3 years ago
Adding and subtracting what is the solution to x = 4 = 12
marshall27 [118]
The answer would be 3 because 3✖️4= 12
8 0
3 years ago
Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered
inn [45]

Answer:

Step-by-step explanation:

2005 AMC 8 Problems/Problem 20

Problem

Alice and Bob play a game involving a circle whose circumference is divided by 12 equally-spaced points. The points are numbered clockwise, from 1 to 12. Both start on point 12. Alice moves clockwise and Bob, counterclockwise. In a turn of the game, Alice moves 5 points clockwise and Bob moves 9 points counterclockwise. The game ends when they stop on the same point. How many turns will this take?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 24$

Solution

Alice moves $5k$ steps and Bob moves $9k$ steps, where $k$ is the turn they are on. Alice and Bob coincide when the number of steps they move collectively, $14k$, is a multiple of $12$. Since this number must be a multiple of $12$, as stated in the previous sentence, $14$ has a factor $2$, $k$ must have a factor of $6$. The smallest number of turns that is a multiple of $6$ is $\boxed{\textbf{(A)}\ 6}$.

See Also

2005 AMC 8 (Problems • Answer Key • Resources)

Preceded by

Problem 19 Followed by

Problem 21

1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

5 0
2 years ago
Integrate this two questions ​
tankabanditka [31]

Simplify the integrands by polynomial division.

\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)

\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)

Now computing the integrals is trivial.

5.

\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}

where we use the power rule,

\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)

and a substitution to integrate the last term,

\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)

8.

\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}

using the same approach as above.

5 0
2 years ago
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