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4 years ago

Will give brainliest answer

Mathematics

1 answer:

Iteru [2.4K]4 years ago

6 0

Answer:

<h2>98 units</h2>

Solution,

Circumference of circle = 615.44 units

Radius = ?

Now,

Circumference of circle = 615.44

$2\pi \: r = 615.44$

$2 \times 3.14 \times r = 615.44$

$6.28r = 615.44$

$r = \frac{615.44}{6.28}$

$r = 98 \: units$

Hope this helps...

Good luck on your assignment...

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How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 6

$\left[\begin{array}{c}8&&6\end{array}\right] = \frac{8!}{(8-6)!6!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} = 28$

When r = 7

$\left[\begin{array}{c}8&&7\end{array}\right] = \frac{8!}{(8-7)!7!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 8

$\left[\begin{array}{c}8&&8\end{array}\right] = \frac{8!}{(8-8)!8!} = \frac{8!}{8!0!} = 1$

The full sequence is: 1,8,28,56,70,56,28,8,1

And the number of terms is 9

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3 years ago

Which ordered pair is included in the solution set to the following system? y > x2 + 3 y < x2 – 3x + 2 (–2, 8) (0, 2) (0,

I am Lyosha [343]

The ordered pair that is a solution of the system is (-2, 8).

<h3>Which ordered pair is included in the solution set to the following system?</h3>

Here we have the system of inequalities:

y > x² + 3

y < x² - 3x + 2

To check which points are solutions of the system, we can just evaluate both inequalities in the given points and see if they are true.

For example, for the first point (-2, 8) if we evaluate it in the two inequalities we get:

8 > (-2)² + 3 = 7

8 < (-2)² - 3*(-2) + 2 = 12

As we can see, both inequalities are true. So we conclude that (-2, 8) is the solution.

(if you use any other of the 3 points you will see that at least one of the inequalities becomes false).

If you want to learn more about inequalities:

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Find two rational expressions a / b and c / d that produce the result x − 1 / x2 when using the following operations. Answers

Mars2501 [29]

Answer:

a) Let $\frac{a}{b}=\frac{-1}{x^2}, \text{ and } \frac{c}{d}=\frac{1}{x}$ .

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Let $\frac{a}{b}=\frac{1}{x}, \text{ and } \frac{c}{d}=\frac{1}{x^2}.$

Observe that

$\frac{a}{b}-\frac{c}{d}=\frac{1}{x}-\frac{1}{x^2}=\frac{x^2-x}{x^3}=\frac{x(x-1)}{xx^2}=\frac{x-1}{x^2}$

Let $\frac{a}{b}=\frac{x-1}{x}, \text{ and } \frac{c}{d}=\frac{1}{x}.$

Observe that

$\frac{a}{b}*\frac{c}{d}=\frac{x-1}{x}*\frac{1}{x}=\frac{(x-1)1}{x*x}=\frac{x-1}{x^2}$

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Step-by-step explanation:

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