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krok68 [10]
3 years ago
14

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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Use the definition of continuity and the properties of limit to show that the function f(x)=x sqrtx/ (x-6)^2 is continuous at x=
jasenka [17]

Answer:

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

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:

  1. The function \\ f(x) exists at x=36.
  2. The limit on both sides of 36 exists.
  3. The value of the function at x=36 is the same as the value of the limit of the function at x = 36.

Therefore:

The value of the function at x = 36 is:

\\ f(36) = \frac{36*\sqrt{36}}{(36-6)^{2}}

\\ f(36) = \frac{36*6}{900} = \frac{6}{25}

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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Answer:

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and $y$ values ranging from $-2$ to $4$ but $-2$ is not included so range is $(-2,4]$

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C is the correct answer
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