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Anna007 [38]
3 years ago
6

10. Write the ratio as a fraction in simplest form. 12 : 42

Mathematics
2 answers:
Pavel [41]3 years ago
8 0

Answer:

2/7

Step-by-step explanation:

12 : 42 is the same as 12/42

simplify the fraction by dividing the numerator and denominator by 6

you get 2/7

Nookie1986 [14]3 years ago
5 0

Answer:

2/9 and 7/9

Step-by-step explanation:

<h2><u>RATIO INTO FRACTION</u></h2><h3>find the common factor for 12 and 42</h3>

12/6 = 2

42/6 = 7

hence ratio simplified is :

2:7

the fraction is :  2 + 7 for the denominator

<u>2/9 and 7/9</u>

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In a trolley ride around an amusement park, a child traveled from one signpost to a second signpost at a constant speed of 125 m
Inessa05 [86]

Answer:

Given speed of trolley = 125 meters per minute

Speed= 125 meters/minute\\\\Speed =\frac{125}{60}meters/second\\\\Speed=2.083m/s

Now we know that

Distance= Speed x time

Thus for the first case since the time of travel is less than 450 seconds thus the distance traveled is less than

Distance < 2.083 x 450 =937.5 meters

hence depending on the given information we cannot come to any conclusion weather the distance travelled is less than 800 m or greater than 800 m.

For the second case

since the time of travel is greater than 400 seconds

Thus the distance traveled is

Distance>2.083\times 400=833.2meters

which is greater than 800 meters.

4 0
3 years 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
3 years ago
F (-2)=x+6. evaluate the function
damaskus [11]
X= -8       that uis your answer for that question :)
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3 years ago
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grin007 [14]
Whats the questions???
4 0
3 years ago
I was just wondering this is one of my questions in my math books and i needed help (Its really long i did all the work i just w
slava [35]

Answer:

no not that wuch

Step-by-step explanation:

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