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pashok25 [27]
3 years ago
12

Please help I need to finish my homework

Mathematics
1 answer:
valina [46]3 years ago
4 0

Answer:

AB= 0.625 units (3 s.f.)

∠BAC= 52.9° (1 d.p.)

∠ABC= 32.1° (1 d.p.)

Step-by-step explanation:

Please see the attached pictures for full solution.

- Find AB using cosine rule

- find ∠BAC using sine rule

- find ∠ABC using angle sum of triangle property

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Prove that.<br><br>lim Vx (Vx+ 1 - Vx) = 1/2 X&gt;00 ​
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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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Answer:

Step-by-step explanation:

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Set the compass slightly larger than half of FG.

Put the point on F and draw a half circle that intersects FG and faces G.

Put the point on G and draw a half circle that intersects FG and faces F.

Draw a line through the intersections of the two half circles, and the intersection of that line with FG is FG's midpoint.

Connect the midpoint with E, and the connecting line is the median.

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Step-by-step explanation:

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