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Advocard [28]
2 years ago
11

Add the following fractionsanswer the all questions.​

Mathematics
1 answer:
ANTONII [103]2 years ago
5 0

Answer:

1a)  1141/660

3/5 + 6/11 +7/12

   find the LCM of the denominators

the LCM of 5,11,12 is 660

after finding the LCM divide it with the denominator.

after dividing 660 by 5 i get 132. So i need to multiply it with 3. Now, i got 396. Do the same with the rest of the numbers.

u will get smth like this: 396+360+385/660

and the final answer will be1141/660.

b) 71/40

c) 211/28

d) 103/20

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Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A
Vlad [161]

Answer:

y(x)=sin(2x)+2cos(2x)

Step-by-step explanation:

y''+4y=0

This is a homogeneous linear equation. So, assume a solution will be proportional to:

e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda

Now, substitute y(x)=e^{\lambda x} into the differential equation:

\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0

Using the characteristic equation:

\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0

Factor out e^{\lambda x}

e^{\lambda x}(\lambda ^2 +4) =0

Where:

e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda

Therefore the zeros must come from the polynomial:

\lambda^2+4 =0

Solving for \lambda:

\lambda =\pm2i

These roots give the next solutions:

y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}

Where c_1 and c_2 are arbitrary constants. Now, the general solution is the sum of the previous solutions:

y(x)=c_1 e^{2ix} +c_2 e^{-2ix}

Using Euler's identity:

e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)

y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\

Redefine:

i(c_1-c_2)=c_1\\\\c_1+c_2=c_2

Since these are arbitrary constants

y(x)=c_1sin(2x)+c_2cos(2x)

Now, let's find its derivative in order to find c_1 and c_2

y'(x)=2c_1 cos(2x)-2c_2sin(2x)

Evaluating    y(0)=2 :

y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2

Evaluating     y'(0)=2 :

y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1

Finally, the solution is given by:

y(x)=sin(2x)+2cos(2x)

5 0
3 years ago
The point (12,5) is on the graph, what do the coordinates tell you about the water in the bucket?
denpristay [2]
I think it’s A but I’m not to sure
4 0
3 years ago
The probability that event M will not occur is 0.8 and the probability that event R will not occur is 0.6. If events M and R can
ololo11 [35]

Answer:

11/25 The right answer is not a part of the options given.

Step-by-step explanation:

Given;

Probability that event M will not occur = 0.8

Probability that event R will not occur = 0.6

Then the ;

Probability that event M will occur = 1 - 0.8 = 0.2

Probability that event R will occur = 1 - 0.6 = 0.4

If events M and R cannot both occur, probability that either event M or event R will occur

=  probability that M occurs and R does not or the probability that R occurs and M does not

= 0.2 × 0.6 + 0.4 × 0.8

= 0.12 + 0.32

= 0.44

= 11/25

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3 years ago
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Vikentia [17]

Answer:

The graph of the equation has roots at x= -b/a and x= -d/c

Step-by-step explanation:

6 0
3 years ago
There are 30 students in an art class. Fourteen of the students are girls.
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Answer: B

Step-by-step explanation: To get the boys part of the ratio you will have to subtract 30 students and 14 students that are girls because all together there are 30 students and since we only know that 14 of them are girls that leaves the rest of the students boys. 30 - 14 = 16  so now we know how many boys there are. 16:14 can be reduced by 2 and that equals to 8:7.

Hope this helps!

8 0
2 years ago
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