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2 years ago

11

Sharon has a carton of quilting notions. She has a total of 52 notions. She has 16 patterns. She has three times as many thimble

s as templates. How many of each item does she have?

1 answer:

alisha [4.7K]2 years ago

7 0

Answer:

well if she has 52 notions and 16 patterns you will have to multiply

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Which of the following are true?

Kitty [74]

Answer: c).

Both options are true. You can see that the whiskers of data set 2 (The lines extending on either side of the box plots) represent a much larger range of data than data set 1, and that the median in data set 2 (the line down the middle of the boxes) is greater than data set 1.

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3 years ago

Substitute y = 2x + 1 y = 3x - 1

Svet_ta [14]

2x+ 1 = 3x - 1 ;
1 + 1 = 3x - 2x ;
2 = x ;
x = 2 ;
Then, y = 2x2 + 1 = 5 ;

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3 years ago

Before leaving to visit Mexico, levant traded 270 american dollars and received 3000 mexican pesos. When he returned from mexico

Makovka662 [10]

Levant will receive about 9 american dollars when he exchanges his pesos.

According to this question, 1 american dollar is worth about 11 pesos. In order to find that, you need to divide 3000/270 which gives you around 11. In order to find out how many dollars he will receive for his pesos, you need to divide 100 by 11 which gives you about 9 dollars.

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3 years ago

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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

The 12 meter tree in front of a building cast a shadow of 5 meters. at the same time the building's shadows is 10 m long how tal

morpeh [17]

Answer:

4.2 i think?

Step-by-step explanation:

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3 years ago

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