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OLEGan [10]
3 years ago
10

It t takes a snail 10 seconds to move 1 centimeter.

Mathematics
2 answers:
tangare [24]2 years ago
5 0

answer is 0.2 using cross multiplication.

swat322 years ago
4 0

Answer:

0.2

Step-by-step explanation:

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You subtract the % that the sale is and take it away from 100%. You get 75% then take the original price and divide it by .75
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Step-by-step explanation:

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2 years ago
John manages 27 apple orchards in North Georgia. He wants to split the orchards into thirds so that he can open 3 local stores.
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Answer: 27 x 1/3

Step-by-step explanation: 27/1 x 1/3 = 9 orchards per store

3 0
1 year ago
Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is
strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is 121y''+110y'-24y=0. We propose that the solution of this equations is of the form y = Ae^{rt}. Then, by replacing the derivatives we get the following

121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

121r^2+110r-24=0

Recall that the roots of a polynomial of the form ax^2+bx+c are given by the formula

x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

r_1 = -\frac{12}{11}

r_2 = \frac{2}{11}

So, in this case, the general solution is y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

c_1 + c_2 = 1

c_1\frac{-12}{11} + c_2\frac{2}{11} = 0(or equivalently c_2 = 6c_1

By replacing the second equation in the first one, we get 7c_1 = 1 which implies that c_1 = \frac{1}{7}, c_2 = \frac{6}{7}.

So y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}

b) By using y(0) =0 and y'(0)=1 we get the equations

c_1+c_2 =0

c_1\frac{-12}{11} + c_2\frac{2}{11} = 1(or equivalently -12c_1+2c_2 = 11

By solving this system, the solution is c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}

Then y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}

c)

The Wronskian of the solutions is calculated as the determinant of the following matrix

\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2

By plugging the values of y_1 and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

e^{\int -p(x) dx}

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}

Note that this function is always positive, and thus, never zero. So y_1, y_2 is a fundamental set of solutions.

8 0
2 years ago
17+(-4)-4/5j-2/5j+6​
Snowcat [4.5K]

Answer:

19 - (6/5)j

Step-by-step explanation:

In this expression we have addition, subtraction and multiplication.

According to order of operations rules, mult. and div. must be carried out before add. or subt.

Therefore, we begin with 17+(-4), which becomes 17 - 4, or 13.

Next we see -4/5j, by which I assume you mean -4/5 * j.

Summarizing what we have so far:

13  - (4/5)j - 2/5j + 6

Combining the j terms, we get

13 - (6/5)j + 6, which collapses to 19 - (6/5)j.

7 0
3 years ago
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