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4 years ago

What is the length of YX?

Mathematics

1 answer:

lozanna [386]4 years ago

4 0

Answer:

15.87 meters.

Step-by-step explanation:

We have been given a diagram of a circle and we are asked to find the length of segment YX.

We can see that line YX is tangent of our given circle and line VZ is secant of our circle.

To find the length of YX we will Tangent-Secant theorem, which states that the product of the lengths of the secant and its external segment equals the square of the length of the tangent segment.

Using Tangent-Secant theorem we can set an equation to solve for YX as:

$(YX)^2=9*(19+9)$

$(YX)^2=9*(28)$

$(YX)^2=252$

Let us take square root of both sides of our equation.

$YX=\sqrt{252}$

$YX=15.874507866\approx 15.87$

Therefore, the length of YX is 15.87 meters.

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3 years ago

Find the difference. (9/x^2-9x)-(6/x^2-81)

Sunny_sXe [5.5K]

The difference is $\frac{3 x+81}{x(x-9)(x+9)}$

Explanation:

The expression is $\left(\frac{9}{x^{2}-9 x}\right)-\left(\frac{6}{x^{2}-81}\right)$

Removing the parenthesis, we have,

$\left\frac{9}{x^{2}-9 x}\right-\left\frac{6}{x^{2}-81}\right$

Factoring the terms $x^{2}-9 x$ and $x^{2}-81$ , we get,

$\frac{9}{x(x-9)}-\frac{6}{(x+9)(x-9)}$

Taking LCM, we get,

$\frac{9(x+9)-6x}{x(x-9)(x+9)}}$

Simplifying the numerator, we get,

$\frac{9x+81-6x}{x(x-9)(x+9)}}$

Subtracting the numerator, we have,

$\frac{3 x+81}{x(x-9)(x+9)}$

Hence, the difference is $\frac{3 x+81}{x(x-9)(x+9)}$

7 0

3 years ago

Can you find the limits of this

Pavel [41]

Answer:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{-3}{8}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Constant]: $\displaystyle \lim_{x \to c} b = b$

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

Limit Property [Addition/Subtraction]: $\displaystyle \lim_{x \to c} [f(x) \pm g(x)] = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

We are given the following limit:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16}$

Let's substitute in x = -2 using the limit rule:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{(-2)^3 + 8}{(-2)^4 - 16}$

Evaluating this, we arrive at an indeterminate form:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \frac{0}{0}$

Since we have an indeterminate form, let's use L'Hopital's Rule. Differentiate both the numerator and denominator respectively:

$\displaystyle \lim_{x \to -2} \frac{x^3 + 8}{x^4 - 16} = \lim_{x \to -2} \frac{3x^2}{4x^3}$

Substitute in x = -2 using the limit rule:

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

Evaluating this, we get:

$\displaystyle \lim_{x \to -2} \frac{3x^2}{4x^3} = \frac{-3}{8}$

And we have our answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

3 years ago