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steposvetlana [31]
3 years ago
10

*URGENT* I don’t understand this and i don’t know the answers, please help!!

Mathematics
1 answer:
motikmotik3 years ago
8 0

Answer:

A) x>8

B) x<4

Step-by-step explanation:

A) 3x+5>29

3x>24

x>8 (It's greater than because the angle opposite is greater than the other angle)

B) 6x-17<7

6x<24

x<4 (It's less than because the angle opposite is smaller than the other angle)

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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
What's the answer to this 94&lt;-2(1+6b)
tigry1 [53]

Answer:

-8 > b

Step-by-step explanation:

94<-2(1+6b)

Distribute

94<-2-12b

Add 2 to each side

94+2<-2-12b +2

96 < -12b

Divide each side by -12, remembering to flip the inequality since we divide by a negative

96/-12 > -12b/-12

-8 > b

5 0
3 years ago
Ty is eight inches shorter than his brother reece. if ty is 42 inches tall, how tall is reece?
Vladimir [108]
Ty is 50in tall
8+42=50
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3 years ago
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MArishka [77]
C and B are correct,
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3 years ago
Read 2 more answers
Form the intersection for the following sets. E = {1, 2, 3} F = {101, 102, 103, 104} E ∩ F =
madam [21]
The intersection of two sets is a set that contains elements that are common to both sets.
There are no common elements in sets E and F, so the answer is the empty set.

<span>E ∩ F =  { }</span>
3 0
3 years ago
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