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cestrela7 [59]
3 years ago
5

Sophia bought 3 yards of trim to put around a rectangular scarf. She wants the width of the scarf to be a whole number that is a

t least 6 inches and at most 12 inches. If she uses all the trim,wh Write your answers in inchesat are the possible dimensions of her scarf?
Mathematics
2 answers:
sergey [27]3 years ago
8 0
Possible dimensions of the scarf are: 
Width:6 inches  Length:96inches 
Width:12 inches Length:84 inches
Width:11inches Length: 86 inches
Width:10 inches Length: 88 inches
Width: 7 inches Length: 94 inches
Width: 8 inches Length: 92 inches
Width: 9 inches Length: 90 inches

Furkat [3]3 years ago
7 0

Answer:

The length of the scarf is at least 42 inches and at most 48 inches.

Step-by-step explanation:

We are given the following information:

Length of trim = 3 yards = 108 inches

The width of the scarf to be a whole number that is at least 6 inches and at most 12 inches.

Let w be the width of the scarf and l be the length of scarf.

Perimeter of scarf =

\text{Permiter of rectangle} = 2(l+w)

Then, we can write:

6 \leq w \leq 12\\6+l \leq w+l \leq 12+l\\2(6+l) \leq 2(l+w) \leq 2(12+l)\\2(6+l) \leq \text{Perimeter of scarf}) \leq 2(12+l)\\2(6+l) \leq 108 \leq 2(12+l)\\\Rightarrow 2(6+l) \leq 108\\12 + 2l \leq 108\\l \leq 48\\\rightarrow 108 \leq 2(12+l)\\108 \leq 24 + 2l\\l \geq 42\\\Rightarrow 42 \leq l \leq 48

The length of the scarf is at least 42 inches and at most 48 inches.

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3 years ago
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Si mi disco duro tiene una capacidad de 1.5 Terabytes ¿ cuántas llaves de 32 Gigabits caben en ese disco duro?
Art [367]

Answer:

The number of 32 Gigabit keys that can be fitted on the hard drive is 375.

Step-by-step explanation:

The question is:

If my hard drive has a capacity of 1.5 Terabytes, how many 32 Gigabit keys can fit on that hard drive?

Solution:

1 Terabyte = 8000 Gigabits

Then 1.5 Terabytes in Gigabits is:

1.5 Terabytes = (8000 × 1.5) Gigabits

                      = 12000 Gigabits

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Compute the number of 32 Gigabit keys that can be fitted on the hard drive as follows:

\text{Number of 32 Gigabit keys}=\frac{12000}{32}=375

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4 0
3 years ago
Help plz math is my worst subject i have a lot of over due work
zubka84 [21]
I believe the answer is the first option. 1 to 3 with intervals of 1.
Hope this helps!

5 0
3 years ago
PLEASE HELP!!!!!!!!!!!!!!!!
Anton [14]

Answer:

<h2>1. subtract 3x, subtract 4, divide by -4</h2><h2 /><h2>2. add x</h2>

Step-by-step explanation:

1 + 3x = -x + 4.

subtract 3x

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subtract 4

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divide by -4

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This is an interesting question. I chose to tackle it using the Law of Cosines.
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Subtracting twice the second equation from the first, we have
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We know that MB = BC/2. When we substitute the given information, we have
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4 0
3 years ago
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