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3 years ago

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Which best describes the diameter of a circle?

2 answers:

kondor19780726 [428]3 years ago

5 0

Which best describes the diameter of a circle?

<em>Statements</em>

<u><em>A. The distance from the center to any point of the circle</em></u>

This statement describes the radius of the circle

<u><em>B. The length of a chord that contains the center of the circle</em></u>

This statement describes the diameter of the circle

<u><em>C. The distance around the circle</em></u>

This statement describes the circumference of the circle

<u><em>D. The length of a chord that does not contain the center of the circle</em></u>

This statement describes a line segment linking any two points on a circle that does not contain the center of the circle

therefore

<u>the answer is the option</u>

B. The length of a chord that contains the center of the circle

enyata [817]3 years ago

3 0

The <em><u>correct answer</u></em> is:

B) The length of a chord that contains the center of the circle

Explanation:

A chord is a segment that goes across a circle, with both endpoints on the circle.

A diameter goes through the center of a circle and has both endpoints on the circle; this makes it a chord that goes through the center.

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Madeline needs to find the slope of the line that passes through the points (9, 12) and (7, 4). She sets up the following work:

Over [174]

Answer:

No

Step-by-step explanation:

She has everything correct but one thing. In order to have her answer correct, she must put the 9 in front of the 7 because the bigger number needs to go in front.

4 0

3 years ago

Which function is an odd function?

hjlf

A function is odd if:
f ( - x ) = - f ( x )
A ) f ( - x )= 0.8 ( - x ) ^3 = - 0.8 x^3 = - f ( x )
Answer: A ) f ( x ) = 0.8 x^3 is an odd function.
Functions B ) and C ) are even and D) is neither even nor odd.

8 0

3 years ago

Two numbers N and 16 have LCM = 48 and GCF = 8. Find N.

RSB [31]

LCM=product of highest occurring primes in the numbers prime factorization.

GCF=product of shared primes in the numbers prime factorization.

16=2*2*2*2

Since the GCF is 8 N and 16 share only 2*2*2

Since the LCM is 48 and 16 has 2*2*2*2 the other number has a factor of 3

So the other number is 2*2*2*3=24

N=24

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

<u>Calculus</u>

Limits

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

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in <em>x</em> = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in <em>x</em> = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in <em>x</em> = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

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

Unit: Limits

6 0

3 years ago

Rewrite each equation, then graph<br> the line<br> 6x + 3y = 12

vekshin1

Here is the answer. I’m not so sure if my answer is correct about which side should it be shaded and whether it’s a dotted or solid line, but I hope you will get some sort of idea .

4 0

3 years ago