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Bingel [31]
2 years ago
7

In a certain board game, a 12-sided number cube

Mathematics
1 answer:
alexandr1967 [171]2 years ago
8 0

Answer:

4/27 ezzzzzzzzzzzzzzzzzzzz

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Which of the following pairs of lines are not parallel?
goldfiish [28.3K]

Answer: Not sure

Step-by-step explanation: A is wrong because they are parallel in one is on the top to the right and 1 is on the bottom. sorry thats all i know :(

7 0
2 years ago
The temperature in the late afternoon
Fantom [35]

Answer:

-7.5 - (8.5 + 5)

Step-by-step explanation:

We just add together the values of how much it dropped at different parts of the day, and subtract them from what we started with. The parenthesis are important cause we want to know what the total temperature drop for the day was before we subtract it to get the actual temperature we end up with.

6 0
3 years ago
Read 2 more answers
PLZ HELP ASAP <br> Find f(x) if it is known that f(x−3)=0.2x−0.6.
iren [92.7K]

Answer:

f(x) = 0.2x

Step-by-step explanation:

hello :

let : f(x) = ax+b    calculate a  and b

f(x-3) =a(x-3)+b = ax+b-3a.....(*)

given    :  f(x-3) = 0.2x-0.6...(**)

by (*) and(**) :  a= 0.2    and   b-3a = - 0.6

b-3(0.2) = - 0.6

so : b=0

conclusion : f(x) = 0.2x

4 0
3 years ago
1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a
katovenus [111]

1. Let a and b be coefficients such that

\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}

\implies 1 = (2a+b)x + 3a

\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23

so that

\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}

2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

so the particular solution to the ODE is

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)

We can solve this explicitly for y :

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)

\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|

\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

8 0
2 years ago
I am very stuck with this
Bingel [31]

Answer:

see explanation

Step-by-step explanation:

To multiply the vector by a scalar, multiply each of the elements by the scalar.

To add 3 vectors add the corresponding elements of each vector

2a + 3b + 4c

= 2\left[\begin{array}{ccc}2\\3\\\end{array}\right] + 3\left[\begin{array}{ccc}-4\\1\\\end{array}\right] + 4\left[\begin{array}{ccc}3\\2\\\end{array}\right]

= \left[\begin{array}{ccc}4\\6\\\end{array}\right] + \left[\begin{array}{ccc}-12\\3\\\end{array}\right] + \left[\begin{array}{ccc}12\\8\\\end{array}\right]

= \left[\begin{array}{ccc}4-12+12\\6+3+8\\\end{array}\right]

= \left[\begin{array}{ccc}4\\17\\\end{array}\right]

8 0
3 years ago
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