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3 years ago

I need help with this equation |5-3x|+2=19

Mathematics

2 answers:

Katen [24]3 years ago

6 0

Answer:

Step-by-step explanation:

STEP

Rearrange this Absolute Value Equation

Absolute value equalitiy entered

5|3x-2|-2 = 19

Another term is moved / added to the right hand side.

5|3x-2| = 21

STEP

Clear the Absolute Value Bars

Clear the absolute-value bars by splitting the equation into its two cases, one for the Positive case and the other for the Negative case.

The Absolute Value term is 5|3x-2|

For the Negative case we'll use -5(3x-2)

For the Positive case we'll use 5(3x-2)

STEP

Solve the Negative Case

-5(3x-2) = 21

Multiply

-15x+10 = 21

Rearrange and Add up

-15x = 11

Divide both sides by 15

-x = (11/15)

Multiply both sides by (-1)

x = -(11/15)

Which is the solution for the Negative Case

STEP

Solve the Positive Case

5(3x-2) = 21

Multiply

15x-10 = 21

Rearrange and Add up

15x = 31

Divide both sides by 15

x = (31/15)

Which is the solution for the Positive Case

STEP

Wrap up the solution

x=-11/15

x=31/15

devlian [24]3 years ago

5 0

Answer:

:) here

Step-by-step explanation:

−

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

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

I need help pls someone

Solnce55 [7]

In this question they've given us the formula and they just want us to plug in the numbers and get the value.

P is the original value, P=$600

r is the depreciation rate, r=10%=0.1

t is the time in years, t=2

$A = P(1-r)^t = 600(1 - 0.1)^2 = 600 \cdot .9^2 = 486$

Answer: $486

7 0

3 years ago

Solve for f (-3) for f (x) = 1 -x<br> f (-3) = [ ? ]

Lostsunrise [7]

F(x) = 1 - x
f(-3) = 1 - (-3)
f(-3) = 4

3 0

3 years ago

Read 2 more answers

Find the quotient of 191 divided by 5

ANEK [815]

38.2 is the quotient!!!!
Hope this helps

8 0

4 years ago

Read 2 more answers

I need help with this equation |5-3x|+2=19​

I need help with this equation |5-3x|+2=19