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Troyanec [42]
3 years ago
13

Please help me with this problem !!!

Mathematics
2 answers:
Alborosie3 years ago
7 0

Answer:

all you have to do is select all numbers that are 0 or less

anything above 0 is wrong

Hope it Helps!

Step-by-step explanation:

Oduvanchick [21]3 years ago
7 0
Anything above 0 is wrong
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0.69465837 is the answer

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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}.

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3 years ago
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An economist read a report that claimed that 60% of residents in her large city owned their homes. She
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Answer:

0.21

Step-by-step explanation:

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on a recent trip together nancy drove 70 miles, dave drove 45 miles and louis drove 95 miles. how many more miles did nancy driv
EleoNora [17]

She drove 25 more miles than Dave.

70 - 45= 25

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The measure of arc LK is?
Oksi-84 [34.3K]

Answer:

Step-by-step explanation:

Because MK is a diameter, then angle L is a right angle. We already know that the measure of angle K is 50, so the measure of angle M has to be 40 because of the triangle angle-sum theorem. The rule for inscribed angles and the arcs they cut off is that the angle is half the measure of its intercepted arc or, likewise, the arc is twice the measure of the angle that cuts it off. Since arc LK is across from angle M and is cut off by angle M, then arc LK is twice the measure of angle M, and is 80. That's the same reason why angle L is 90; arc MK is a semi-circle, with a degree measure of 180, and angle L is half of that.

Arc LK = 80

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3 years ago
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