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KATRIN_1 [288]
3 years ago
7

What is the value of x in the equation –x = 4 – 3x + 6?

Mathematics
2 answers:
Nataly [62]3 years ago
7 0
Remember you can do anything to an equaiton as long as you do it to both sides


-x=4-3x
add 3x both sides
3x-x=4+3x-3x
2x=4+0
2x=4
divide 2
2x/2=4/2
2/2x=2
1x=2
x=2
Dima020 [189]3 years ago
7 0

Answer:

The answer is x=5

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The answer is 0.40 or 0.4
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Find the solution of w(-15-w)=0
inna [77]

Answer:

w(-15-w)= 0

w= 0 or (-15-w) = 0

Now,

-15-w =0

w= -15

So,The solution of w(-15-w) = 0

w= 0, -15

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Uber charges a $5 pickup fee and $0.25 per mile. define variables and write an equation to model this situation.
Degger [83]

Answer:

See below

Step-by-step explanation:

<h3>Uber</h3>
  • $5 pickup fee
  • $0.25 per mile

<u>Equation</u>

  • y = 0.25x + 5

<u>Cost of 5 miles:</u>

  • y = 0.25(5) + 5 = $6.25
<h3>KC yellow Cab Co.</h3>
  • $1.75 pickup fee
  • $0.75 per mile

<u>Equation</u>

  • y = 0.75x + 1.75

<u>Cost of 5 miles:</u>

  • y = 0.75(5) + 1.75 = $5.5

Comparing the cost we see KC yellow Cab Co. is better value for a 5 mile ride.

8 0
3 years ago
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)

3 0
3 years ago
A bus company is replacing all the old seats in its buses with new ones. The company owns 15 buses and each bus has 22 seats. It
Pachacha [2.7K]

Answer:

30112.50. 15x22x91.20+26.50

3 0
2 years ago
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