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3 years ago

Point C on line segment BD. Given BD = 5x, BC = 4x, and CD = 4,

Mathematics

1 answer:

Llana [10]3 years ago

4 0

Answer:

The numerical length of BD is 20 units

Step-by-step explanation:

Let us solve the question

∵ Point C lies on the line segment BD

→ That means C divides BD into 2 parts

∴ C divides BD into 2 parts BC and CD

∴ BD = BC + CD

∵ BD = 5x

∵ BC = 4x

∵ CD = 4

→ Substitute them in the equation above

∴ 5x = 4x + 4

→ Subtract 4x from both sides

∵ 5x - 4x = 4x - 4x + 4

∴ x = 4

→ Substitute the value of x in the expression of BD

∵ BD = 5x

∵ x = 4

∴ BD = 5(4) = 20

∴ The numerical length of BD is 20 units

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3 years ago

Plz with steps plzzzzzz

Stella [2.4K]

Answer: $-\frac{\sqrt{2a}}{8a}$

=======================================================

Explanation:

The (x-a) in the denominator causes a problem if we tried to simply directly substitute in x = a. This is because we get a division by zero error.

The trick often used for problems like this is to rationalize the numerator as shown in the steps below.

$\displaystyle \lim_{x\to a} \frac{\sqrt{3a-x}-\sqrt{x+a}}{4(x-a)}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x}-\sqrt{x+a})(\sqrt{3a-x}+\sqrt{x+a})}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{(\sqrt{3a-x})^2-(\sqrt{x+a})^2}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-(x+a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{3a-x-x-a}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\$

$\displaystyle \lim_{x\to a} \frac{2a-2x}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(-a+x)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2(x-a)}{4(x-a)(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\\lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\$

At this point, the (x-a) in the denominator has been canceled out. We can now plug in x = a to see what happens

$\displaystyle L = \lim_{x\to a} \frac{-2}{4(\sqrt{3a-x}+\sqrt{x+a})}\\\\\\L = \frac{-2}{4(\sqrt{3a-a}+\sqrt{a+a})}\\\\\\L = \frac{-2}{4(\sqrt{2a}+\sqrt{2a})}\\\\\\L = \frac{-2}{4(2\sqrt{2a})}\\\\\\L = \frac{-2}{8\sqrt{2a}}\\\\\\L = \frac{-1}{4\sqrt{2a}}\\\\\\L = \frac{-1*\sqrt{2a}}{4\sqrt{2a}*\sqrt{2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{2a*2a}}\\\\\\L = \frac{-\sqrt{2a}}{4\sqrt{(2a)^2}}\\\\\\L = \frac{-\sqrt{2a}}{4*2a}\\\\\\L = -\frac{\sqrt{2a}}{8a}\\\\\\$

There's not much else to say from here since we don't know the value of 'a'. So we can stop here.

Therefore,

$\displaystyle \lim_{x\to a} \frac{\sqrt{3a-x}-\sqrt{x+a}}{4(x-a)} = -\frac{\sqrt{2a}}{8a}\\\\\\$

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