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Mademuasel [1]
3 years ago
5

100 POINTS AND BRAINLIEST! SUPPER EASY IM JUST DU.MB!

Mathematics
2 answers:
NikAS [45]3 years ago
7 0

"inspiring that a person can say we own ourselves and no other person does"

Step-by-step explanation:

Svet_ta [14]3 years ago
4 0

Answer:

7hrs late but uhm

Locke is relating to slavery, specifically saying it isn’t right to own someone else or like control them.

Step-by-step explanation:

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Marty’s bike ride is represented by the graph shown on the right.
Anastasy [175]

Answer:

10 m/h

Step-by-step explanation:

By finding the slope of the line you find how fast he is traveling

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Convert the angle −4.5 radians to degrees, rounding to the nearest 10th
olya-2409 [2.1K]
-4.5 rad = -4.5 x 180 / pi = -257.83°
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Which is the equation of a line that has a slope of 1/2 and passes through point 0 , -2
S_A_V [24]
The general equation of a line is y = mx + b
m is given as 1/2, so we have:
y=\frac{1}{2}x+b
Plugging in the given values of x and y, we get:
-2 = 0 + b
Therefore b = -2, and the answer is:
y=\frac{1}{2}x-2 
3 0
4 years ago
(+1) + (-5) - (+5) + (+9)
olchik [2.2K]

Answer:

The answer is 0

Hope that helps!

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2 years ago
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Prove that.<br><br>lim Vx (Vx+ 1 - Vx) = 1/2 X&gt;00 ​
faltersainse [42]

Answer:

The idea is to transform the expression by multiplying (\sqrt{x + 1} - \sqrt{x}) with its conjugate, (\sqrt{x + 1} + \sqrt{x}).

Step-by-step explanation:

For any real number a and b, (a + b)\, (a - b) = a^{2} - b^{2}.

The factor (\sqrt{x + 1} - \sqrt{x}) is irrational. However, when multiplied with its square root conjugate (\sqrt{x + 1} + \sqrt{x}), the product would become rational:

\begin{aligned} & (\sqrt{x + 1} - \sqrt{x}) \, (\sqrt{x + 1} + \sqrt{x}) \\ &= (\sqrt{x + 1})^{2} -(\sqrt{x})^{2} \\ &= (x + 1) - (x) = 1\end{aligned}.

The idea is to multiply \sqrt{x}\, (\sqrt{x + 1} - \sqrt{x}) by \displaystyle \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}} so as to make it easier to take the limit.

Since \displaystyle \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}} = 1, multiplying the expression by this fraction would not change the value of the original expression.

\begin{aligned} & \lim\limits_{x \to \infty} \sqrt{x} \, (\sqrt{x + 1} - \sqrt{x}) \\ &= \lim\limits_{x \to \infty} \left[\sqrt{x} \, (\sqrt{x + 1} - \sqrt{x})\cdot \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}}\right] \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}\, ((x + 1) - x)}{\sqrt{x + 1} + \sqrt{x}} \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + 1}+ \sqrt{x}}\end{aligned}.

The order of x in both the numerator and the denominator are now both (1/2). Hence, dividing both the numerator and the denominator by x^{(1/2)} (same as \sqrt{x}) would ensure that all but the constant terms would approach 0 under this limit:

\begin{aligned} & \lim\limits_{x \to \infty} \sqrt{x} \, (\sqrt{x + 1} - \sqrt{x}) \\ &= \cdots\\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + 1}+ \sqrt{x}} \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x} / \sqrt{x}}{(\sqrt{x + 1} / \sqrt{x}) + (\sqrt{x} / \sqrt{x})} \\ &= \lim\limits_{x \to \infty}\frac{1}{\sqrt{(x / x) + (1 / x)} + 1} \\ &= \lim\limits_{x \to \infty} \frac{1}{\sqrt{1 + (1/x)} + 1}\end{aligned}.

By continuity:

\begin{aligned} & \lim\limits_{x \to \infty} \sqrt{x} \, (\sqrt{x + 1} - \sqrt{x}) \\ &= \cdots\\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + 1}+ \sqrt{x}} \\ &= \cdots \\ &= \lim\limits_{x \to \infty} \frac{1}{\sqrt{1 + (1/x)} + 1} \\ &= \frac{1}{\sqrt{1 + \lim\limits_{x \to \infty}(1/x)} + 1} \\ &= \frac{1}{1 + 1} \\ &= \frac{1}{2}\end{aligned}.

8 0
3 years ago
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