How many solutions does 6 - 3x = 4 - x - 3 - 2x have?

Mathematics

1 answer:

White raven [17]3 years ago

7 0

Answer:

No Solution

Step-by-step explanation:

6 - 3x = 4 - x - 3 - 2x (Given)

6 - 3x = -1 -3x (combine like terms)

7 (addition)

The x cancels each other out so no solution.

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Evaluate : limx→ tan2-sin2x x3

GrogVix [38]

Given:

$\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}$

Solve:

$\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}$

Use l'hopital's rule:

$\begin{gathered} =\lim _{x\to0}\frac{\frac{d}{dx}(-\sin 2x+\tan 2x)}{\frac{d}{dx}(x^3)} \\ =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \end{gathered}$

Simplify:

$\begin{gathered} =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \\ =\lim _{x\to0}\frac{2(-\cos (2x)+\tan ^2(2x)+1)}{3x^2} \end{gathered}$

Apply the constant multiple rule:

$\begin{gathered} \lim _{x\to0}cf(x)=c\lim _{x\to0}f(x) \\ \text{With c=}\frac{2}{3} \\ f(x)=\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2} \end{gathered}$ $\begin{gathered} =\frac{2\lim _{x\to0}\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2}}{3} \\ =\frac{2\lim _{x\rightarrow0}\frac{(4\tan ^2(2x)+4)\tan (2x)+2\sin (2x)}{2x}}{3} \end{gathered}$

Similary :

$\begin{gathered} =\frac{2\lim _{x\to0}(2\cos (2x)+12\tan ^4(2x)+16\tan ^2(2x)+4)}{3} \\ =\frac{2(6)}{3} \\ =4 \end{gathered}$

8 0

1 year ago

Can you please help me i’ll give extra points

Sauron [17]

First, we need to solve for sin(67°) = 22/x. Multiply each side by x and divide each side by sin(67°)

x = 22/sin(67°) = 23.9

Use pythagorean theorem to find the last side.

a^2 + 22^2 = 23.9^2

a^2 + 484 = 571.21. Subtract each side by 484

a^2= 87.21. Take the square root of each side.

a = 9.3

x = 23.9 and the other side is 9.3

4 0

3 years ago

The production manager is going to present this information to the company's board of directors. Which graph should the manager

viktelen [127]

To best emphasize the number of defects. Manager should use graph 3 (refer the image shown):

If we talk about graph 1, it can also be used but usually we put the time line on the horizontal axis, for the convenience and the quantity to be measured on the y-axis. In the graph 1, the time is placed on the vertical axis (x-axis) so it would not be a good pick for the manager.

Same is the case with graph 2 again we have time on the vertical axis. So it is not a good idea to with graph 2.

Graph 3 could be the best to emphasize the number of defects because first of all time is placed on the horizontal axis and the quantity to be shown is on the vertical axis. Secondly, the range of the vertical axis is less so it is easy to observe the data set on the graph quite distinctively. Therefore, graph 3 is the best pick.

Graph 4 is placed correctly in terms of vertical and horizontal axes but the range of vertical axis is quite high due to which the dispersion or the display of the data is quite compressed and it gets hard to visualize.