X=1, X^2=X, X^2-1=X-1, (X+1)(X-1)=X-1, X+1=1, X=0 Where is the mastic?

Mathematics

1 answer:

Alchen [17]3 years ago

5 0

Answer:

You are technically dividing by zero, it breaks math.

In the step (X+1)(X-1)=X-1 → X+1=1.

You divide (X+1)(X-1)=X-1 by (x-1) on both sides to get:

(x+1)(x-1) / (x-1) = (x-1) / (x-1).

Since x = 1,

(x + 1)(x - 1) / (x - 1) = (x - 1) / (x - 1)

becomes (2)(0) / 0 = 0 / 0.

And when you divide by zero, it's undefined since you cannot multiply a number by zero and get a number other than zero, it's undefinable.

This is where limits come in, when you divide by a variable, you have to assumed that is it ≠ 0.

You know that 1 ≠ 2, and 0 ≠ 1, and so on.

As dividing by zero approaches infinity,

1/0.1 = 10, 1/0.01 = 100, 1/0.001 = 1000, 1/0.00..1 = 1000.. → 1/0 → infinity.

1/-0.1 = -10, 1/-0.01 = -100, 1/-0.001 = -1000,

1/-0.00..1 = -1000.. → 1/0 → -infinity.

This suggests that dividing by zero should be infinity.

Therefore infinity = -infinity, Wrong.

When you are dividing, you are really asking yourself, how many times does this number fit into the other number.

1 cannot fit into 0, there isn't anything to fit into. This can also be thought of as:

1 = 0 + 0 + 0 + 0 + 0 + ......

0 + 0 = 0, so 1 ≠ 0.

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Answer:

Step-by-step explanation:

The model fo the shell is given by the following equation of equilibrium:

$\Sigma F = -b\cdot v^{2} - m\cdot g = m\cdot \frac{dv}{dt}$

This first-order differential equation has separable variables, which are cleared herein:

$\int\limits^t_{0\,s} \, dt = -\frac{m}{b} \int\limits^{0\,\frac{m}{s} }_{600\,\frac{m}{s} } {\frac{1}{ v^{2}+\frac{m}{b}\cdot g } } \, dv$

The solution of this integral is:

$t = -\frac{m}{2b}\cdot \left[\tan^{-1} \left(\frac{v}{\sqrt{\frac{m\cdot g}{b} } }\right) - \tan^{-1} \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } }\right)\right]$

$\tan^{-1} \left(\frac{v}{\sqrt{\frac{m\cdot g}{b} } } \right)=-\frac{2\cdot b\cdot t}{m} + \tan^{-1}\left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)$

$\frac{v}{\sqrt{\frac{m\cdot g}{b} } }=\tan \left[-\frac{2\cdot b\cdot t}{m} + \tan^{-1}\left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)\right]$

$v = \sqrt{\frac{m\cdot g}{b} } \left [\frac{\tan \left(-\frac{2\cdot b \cdot t}{m} \right)+ \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)}{1 - \left(\frac{600}{\sqrt{\frac{m\cdot g}{b} } } \right)\cdot \tan \left(-\frac{2\cdot b \cdot t}{m} \right) }\right]$

4 0

3 years ago

Jarvis left out a statement from the proof shown.which statement is best explained by the corresponding angles postulate

Salsk061 [2.6K]

The postulate of the corresponding angles establishes that when a transversal line cuts two parallel lines, the corresponding angles are congruent. These angles are on the same side of the parallel lines and on the same side of the transversal line.

Then, if we based on this definition and analize the figure attached, we can notice that the angles ∠1 and ∠3 are corresponding angles, so they are congruent. In this case the angle ∠1 is internal and the angle ∠3 is external.

The answer is: ∠1 and ∠3 are congruent (See the image attached).