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Nady [450]
3 years ago
6

Consider a circle whose size can vary. Let r represent the radius of the circle (in cm) and let C represent the circumference of

the circle (in cm). Suppose the function f determines the circumference of the circle in cm, C , given its radius length in cm, r .'
a. Write a function formula for f.

b. What does f(12) represent in this context? Select all that apply.

i. The radius (in cm) of a circle whose circumference is 12 cm.
ii. A circle with a radius of 12 cm.
iii. The circumference (in cm) of a circle whose radius is 12 cm.
iv. A circle with a circumference of 12 cm.

c. If f(a) 12, what does a represent in this context? Select all that apply.
i. The circumference (in cm) of a circle whose radius is 12 cm.
ii. The radius (in cm) of a circle whose circumference is 12 cm.
iii. A circle with a circumference of 12 cm.
iv. A circle with a radius of 12 crm

d. What values can r assume in this context?
e. What values can f(r) assume in this context?
Mathematics
1 answer:
lora16 [44]3 years ago
8 0

Answer and Step by Step Explanation

a) f(r) = 2πr

The question only asks for the expression for circumference in which the value of the radius can always be plugged in and a circumference obtained. And this, from basic geometry is 2πr.

b) Option (iii) is correct.

f(12) represents the circumference (in cm) of a circle whose radius is 12 cm.

A radius of r = 12 cm was plugged into the circumference equation and f(12) is the answer obtained.

c) Option (ii) is correct.

In the context f(a) = 12, a represents the radius (in cm) of a circle whose circumference is 12 cm

A radius of r = a cm was plugged into the circumference equation and 12 is the answer for circumference obtained.

d) In this context, the radius can take on values of every positive real number.

This is because simply, the radius of a circle is defined as a straight line from the center of a circle to the circumference of a circle.

Although, generally, a radius can take on virtually any real or imaginary number because more technically, a circle is the locus of all points equidistant from a central point. And this opens the scope of the radius. The general equation of a circle with centre (a,b) and radius, r is given as

(x - a)² + (y - b)² = r²

The square on the r, gives it the power to have such a wide range of values.

e) In this context, the circumference can take on values of every positive real number.

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The rocket hits the ground at a time of 11.59 seconds.

Step-by-step explanation:

The height of the rocket, after x seconds, is given by the following equation:

y = -16x^2 + 177x + 98

It hits the ground when y = 0, so we have to find x for which y = 0, which is a quadratic equation.

Finding the roots of a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

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In this question:

y = -16x^2 + 177x + 98

-16x^2 + 177x + 98 = 0

So

a = -16, b = 177, c = 98

\bigtriangleup = 177^{2} - 4(-16)(98) = 37601

x_{1} = \frac{-177 + \sqrt{37601}}{2*(-16)} = -0.53

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Total number of contestants at the beginning = 50,000

After first day, 20% 0f 50,000 are removed.

That is, (20/100)×50,000 = 10,000

Contestants remaining after first day = 50,000-10,000 = 40,000

Second day:

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That is (20/100)×40000=8000

Balance =40000-8000=32,000

Third day:

Contestants= 32000

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That is (20/100)×32000=6400

Balance =32000-6400=25,600

Fouth day:

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Fifth day:

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Sixth day:

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That is (20/100)×13,107.2=2621.44

Balance =13107.2-2621.44=10485.76

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Step-by-step explanation:

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