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3 years ago

Help me pls, BRAINEST AWARD

Mathematics

1 answer:

Natali [406]3 years ago

4 0

Answer:

x = 3.7

Step-by-step explanation:

By applying sine ratio for the given angle B,

sin(39°) = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

sin(39°) = $\frac{AD}{AB}$

0.6293 = $\frac{AD}{7}$

AD = 4.41

By applying tangent ratio for the given angle C,

tan(50°) = $\frac{\text{Opposite side}}{\text{Adjacent side}}$

1.19 = $\frac{AD}{x}$

1.19 = $\frac{4.41}{x}$

x = 3.7

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Answer:

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Step-by-step explanation:

We need to follow the following steps:

The function is:

$\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$

The function is continuous at point x=36 if:

The function $\\ f(x)$ exists at x=36.
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The value of the function at x=36 is the same as the value of the limit of the function at x = 36.

Therefore:

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The limit of the $\\ f(x)$ is the same at both sides of x=36, that is, the evaluation of the limit for values coming below x = 36, or 33, 34, 35.5, 35.9, 35.99999 is the same that the limit for values coming above x = 36, or 38, 37, 36.5, 36.1, 36.01, 36.001, 36.0001, etc.

For this case:

$\\ lim_{x \to 36} f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$

$\\ \lim_{x \to 36} f(x) = \frac{6}{25}$

Since

$\\ f(36) = \frac{6}{25}$

And

$\\ \lim_{x \to 36} f(x) = \frac{6}{25}$

Then, the function $\\ f(x) = \frac{x*\sqrt{x}}{(x-6)^{2}}$ is continuous at x = 36.

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