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GalinKa [24]
3 years ago
13

What are the values of x in the equation shown 2|3x-4|=20

Mathematics
2 answers:
ipn [44]3 years ago
6 0

Answer:14/3

Step-by-step explanation:

2(3x-4)=20

Divide both sides by2

2(3x-4)/2=20/2

3x-4=10

Add 4 to both sides

3x-4+4=10+4

3x=14

Divide both sides by 3

3x/3=14/3

x=14/3

Vlada [557]3 years ago
3 0
2(3x-4)=20
Divide both side by 2
3x-4=10
Solve the equations
x= 14/3
Or
x=-2
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Evaluate: 542 × 31
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Step-by-step explanation:

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3 years ago
Kaylee invested $910 in an account paying an interest rate of 2.6% compounded
Nikitich [7]

Answer:

$1,179

Step-by-step explanation:

Lets use the compound interest formula provided to solve this:

A=P(1+\frac{r}{n} )^{nt}

<em>P = initial balance</em>

<em>r = interest rate (decimal)</em>

<em>n = number of times compounded annually</em>

<em>t = time</em>

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First, lets change 2.6% into a decimal:

2.6% -> \frac{2.6}{100} -> 0.026

Since the interest is compounded quarterly, we will use 4 for n. Lets plug in the values now:

A=910(1+\frac{0.026}{4})^{4(10)}

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The account balance after 10 years will be $1,179

8 0
3 years ago
Which equation shows how (-10,8) can be used to write the equation of this line in point-slope from?
NeTakaya

Options :

y – 8 = –0.15(x – 10)

y + 8 = –0.15(x – 10)

y – 8 = –0.2(x + 10)

y + 8 = –0.2(x – 10)

Answer:

y – 8 = –0.2(x + 10)

Step-by-step explanation:

(-10, 8)

General equation in point - slope form:

y - y1 = m(x - x1)

m = slope

Given the points : ( - 10, 8)

x1= - 10 ; y1 = 8

y - y1 = m(x - x1)

Substitute the values of x and y into the equation :

y - 8 = m(x - - 10)

y - 8 = m(x + 10)

The best option which fits is :

y – 8 = –0.2(x + 10)

6 0
3 years ago
Read 2 more answers
x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco
igomit [66]

Answer:

x=-cos(t)+2sin(t)

Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0

Will have a characteristic equation of the form:

a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0

Where solutions r_1,r_2...,r_n are the roots from which the general solution can be found.

For real roots the solution is given by:

y(t)=c_1e^{r_1t} +c_2e^{r_2t}

For real repeated roots the solution is given by:

y(t)=c_1e^{rt} +c_2te^{rt}

For complex roots the solution is given by:

y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)

Where:

r_1_,_2=\lambda \pm \mu i

Let's find the solution for x''+x=0 using the previous information:

The characteristic equation is:

r^{2} +1=0

So, the roots are given by:

r_1_,_2=0\pm \sqrt{-1} =\pm i

Therefore, the solution is:

x(t)=c_1cos(t)+c_2sin(t)

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of x(t) in order to find the constants c_1 and c_2:

x'(t)=-c_1sin(t)+c_2cos(t)

Evaluating the initial conditions:

x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1

And

x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2

Now we have found the value of the constants, the solution of the second-order IVP is:

x=-cos(t)+2sin(t)

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3 years ago
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