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2 years ago

YOUR TURN

Mathematics

1 answer:

Ivahew [28]2 years ago

8 0

Answer:

$Width: \frac{3in}{8ft} = \frac{5in}{x} (cross multiply)$

= $40$ ÷3 ≈ 13.3ft

$Width: \frac {3in}{8ft} = \frac{6.5in}{x}$

= 17.3

$A=e$ × $w$

$= 13.3$ × $17.3$ ≈ $230.09$ ≈ $230.1$

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Jose earned 15 points in a video game. He lost 40 points, earned 87 points, then lost 30 more points. Write and evaluate an expr

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(15-40)+(87-30)

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2 years ago

Factor the following expression completely. 16x^5-x^3 A. B. C. D.

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Answer:

x^3(2x+1)(2x-1)

Step-by-step explanation:

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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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Lana71 [14]

Answer:

$36^{6} -26^6$

Step-by-step explanation:

We have 36 characters in total which we can use in order to form a 6 character password.

For each place in this password we have choice of 36 of them, so, the total number of possible passwords of 6 characters is:

$36*36*36*36*36*36=36^6$

But, we have to exclude all passwords which have no numbers because, by the task, the password must contain at least one digit.

So, now we have choice of 26 characteristic to form password without digits and get $26^6$ passwords like that, contained just of letters without digits.

So, the final answer is that we have $36^6-26^6$ 6-character passwords with at least one digit.

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