Resolver (x-1)(-1) | Microsoften solucionagailu matematikoa

(x - 1) (- 1)

Ebaluatu

1 - x

Zabaldu

1 - x

Grafikoa

Azterketa

Polynomial

(x - 1) (- 1)

Bilaketaren antzeko arazoak webgunean

Find the first-variational curve which corresponds to the functional

\int_{- 1}^{1} t^{2} \overset{x}{˙}^{2} d t

when

x (- 1) = - 1

and

x (1) = 1

https://math.stackexchange.com/questions/1023944/find-the-first-variational-curve-which-corresponds-to-the-functional-int-1/1023998

You are wrong here:

\int_{- 1}^{1} 2 t^{2} \overset{x}{˙} \dot{h} d t = 2 \overset{x}{˙} \dot{h} [\frac{1}{3} t^{3}]_{t = - 1}^{t = 1}

Note

x (t)

h (t)

are functions, you cannot pull out them from the integral. ...

How do you solve and graph

x - 1 < - 7

https://socratic.org/questions/how-do-you-solve-and-graph-x-1-7

x < - 6

Explanation: You can arrange your equation

x - 1 + 1 < - 7 + 1

x < - 6

Graph is provided below This is your condition. ...

How do you solve

2 x - 3 > 6

x - 1 < - 8

https://socratic.org/questions/how-do-you-solve-2x-3-6-or-x-1-8

x > 4.5

x < - 7

Explanation: For the first inequality, add 3 to both sides to give:

2 x > 9

and then dividing both sides by

2

...

Factoring Polynomial using Quadratic Equation

https://math.stackexchange.com/questions/1882350/factoring-polynomial-using-quadratic-equation

Look at the expansion

(x + a)^{2} = x^{2} + 2 a x + a^{2} .

Notice that the coefficient of

x^{2}

is always

1

. Thus, if a quadratic polynomial can be written in this form, it must have a leading coefficient of ...

Absolute Value inequality help:

∣ x + 1 ∣ \geq 3

https://math.stackexchange.com/questions/549429/absolute-value-inequality-help-x1-geq-3/549441

The difference

∣ a - b ∣

signifies the positive distance between two numbers

a, b

on the real line. So when you consider

∣ x + 1 ∣

, this doesn't mean the distance between

x

and

1

, because it isn't a ...

Finding partial derivative plug in a value?

https://math.stackexchange.com/questions/703583/finding-partial-derivative-plug-in-a-value

For

F_{x}

, we have \begin{align} F_x &= \frac{\partial}{\partial x} (x^2 \tan^{-1}\frac{x}{y})\\ &= \frac{\partial}{\partial x} (x^2) \cdot \tan^{-1} \frac{x}{y} + x^2 \cdot \frac{\partial}{\partial ...

Partekatu

Kopiatu portapapeletan

-x+1

Erabili banaketa-propietatea x-1 eta -1 biderkatzeko.

-x+1

Erabili banaketa-propietatea x-1 eta -1 biderkatzeko.

Adibideak

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Mugak

x \to - 3 lim \frac{x ^{2} - 9}{x ^{2} + 2 x - 3}

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Gaiak

Bilaketaren antzeko arazoak webgunean

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