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Rufina [12.5K]
3 years ago
9

How do you simplify the mathematical expression: 4(x-6) +3x

Mathematics
2 answers:
nikitadnepr [17]3 years ago
5 0
First you would distribute the 4 to the x and the -6. So the equation would then look like this 4x-24+3x. Next you would combine like terms which would be 4x and 3x. So it would now be 7x-24. You can't simplify it anymore.
Juli2301 [7.4K]3 years ago
4 0
The simplification is x * 7 - 24.
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Anyone!!! Please need this asap
coldgirl [10]

Answer:

angle 3 = angle 6

angle 1 = angle 5

angle 2 + angle 5 + angle 6 = 180°

8 0
4 years ago
Read 2 more answers
Given mſn, find the value of x.<br> (6x-7)º<br> m<br> (4x-10°
Pavel [41]

Answer:

x = 3

Step-by-step explanation:

These angles are actually equal to each other

This is because when two lines intersect, the two opposite angles are the same.

Since they are the same, you can set them equal to each other

Like so:

6x - 7 = 4x - 1

Then solve:

6x - 7 = 4x - 1

6x = 4x +6

2x = 6

x = 3

5 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
Two ropes, AD and BD, are tied to a peg on the ground at point D. The other ends of the ropes are tied to points A and B on a fl
Arlecino [84]

Answer:

  A.  3.66 feet

Step-by-step explanation:

Using trigonometry, we know that ...

  BC = DC·tan(30°)

  AC = DC·tan(45°)

  AB = AC -BC = DC(tan(45°) -tan(30°))

  AB = 5√3·(1 -√3/3) = 5(√3 -1)

  AB ≈ 3.66 . . . feet

8 0
3 years ago
Number 7 <br> solve for c
s2008m [1.1K]

Answer:

60

Step-by-step explanation:

Since equalateral. The angles are the same.

4 0
3 years ago
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