What is the answer to this equation-5=X+X-5

Mathematics

1 answer:

Goryan [66]3 years ago

5 0

I would think x is equal to 0, since 0-5=-5, and of course 0+0=0

You might be interested in

-6x-y=11<br> What’s the answer?

Serhud [2]

Answer:

_6x-y=11/ 6x+y+11=0 this is the answer

3 0

3 years ago

Read 2 more answers

Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

7 0

3 years ago

Find the solutions of the quadratic equation -9c² +2 +3 = 0.

Tresset [83]

Given that,

An equation : -9c² +2c +3 = 0

To find,

Find the value of x.

Solution,

We have, -9c² +2c +3 = 0

We can solve it using the formula as follows :

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, a = -9, b = 2 and c =3

Put the values,

$c=\dfrac{-2\pm \sqrt{(2)^2-4(-9)(3)}}{2(-9)}\\\\c=\dfrac{-2+\sqrt{(2)^{2}-4(-9)(3)}}{2(-9)}, \dfrac{-2-\sqrt{(2)^{2}-4(-9)(3)}}{2(-9)}\\\\c=-0.47, 0.69$