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DochEvi [55]
3 years ago
6

You are wrapping a gift box. You want to tie a ribbon around the box , so that whenever the ribbon intersects an edge of the box

, its distance to the nearest corner is 1/3 the length of the box. How long must the ribbon be?

Mathematics
1 answer:
Paha777 [63]3 years ago
7 0

You are wrapping a gift box. The length of ribbon = length + width + height of box.

<u>Solution:</u>

Given, You are wrapping a gift box.  

You want to tie a ribbon around the box, so that whenever the ribbon intersects an edge of the box, its distance to the nearest corner is 1/3 the length of the box.  

We have to find how long must the ribbon be?

The figure of this problem is attached below

Now, assume we started to tie at one edge, then next we can tie at straight parallel edge, we can continue till our process ends.

Then, total length of the ribbon = length of box + width of box + height of box.

Hence, the length of ribbon = length + width + height of box.

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What is the intersection of the sets A = {2, 5, 6, 14, 16} and B = {1, 3, 6, 8, 14}?
djyliett [7]

Answer:

my brain

Step-by-step explanation:

i cant

what the hell even IS this!

here take this instead.

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now I need a Tylenol, u gave me a migraine.

8 0
3 years ago
Huddson has practiced a 360-back flip for months. This week, he did it perfectly 24 out of 54 times. At this rate, how many time
iris [78.8K]
Let x = perfect flips out of nine
\frac{24}{54}  =  \frac{x}{9}
54x = 216
x = 4
4 flips
5 0
3 years ago
Which of the following is equivalent to tan2θcos(2θ) for all values of θ for which tan2θcos(2θ) is defined?
Aloiza [94]

Answer:

2sin²θ - tan²θ

Step-by-step explanation:

Given

tan²θcos(2θ)

Required

Simplify

We start by simplifying cos(2θ)

cos(2θ) = cos(θ+θ)

From Cosine formula

cos(A+A) = cosAcosA - sinAsinA

cos(A+A) = cos²A - sin²A

By comparison

cos(2θ) = cos(θ+θ)

cos(2θ) = cos²θ - sin²θ ----- equation 1

Recall that cos²θ + sin²θ = 1

Make sin²θ the subject of formula

sin²θ = 1 - cos²θ

Substitute sin²θ = 1 - cos²θ in equation 1

cos(2θ) = cos²θ - (1 - cos²θ)

cos(2θ) = cos²θ - 1 +cos²θ

cos(2θ) = cos²θ + cos²θ - 1

cos(2θ) = 2cos²θ - 1

Substitute 2cos²θ - 1 for cos(2θ) in the given question

tan²θcos(2θ) becomes

tan²θ(2cos²θ - 1)

Open brackets

2cos²θtan²θ - tan²θ

------------------------

Simplify tan²θ

tan²θ = (tanθ)²

Recall that tanθ =  sinθ/cosθ

So, we have

tan²θ = (sinθ/cosθ)²

tan²θ = sin²θ/cos²θ

------------------------

Substitute sin²θ/cos²θ for tan²θ

2cos²θtan²θ - tan²θ becomes

2cos²θ(sin²θ/cos²θ) - tan²θ

Open bracket (cos²θ will cancel out cos²θ) to give

2(sin²θ) - tan²θ

2sin²θ - tan²θ

Hence, the simplification of tan²θcos(2θ) is 2sin²θ - tan²θ

Option E is correct

7 0
3 years ago
I need help on this problem..please and thanks u
GarryVolchara [31]

Answer:

A is on a line with the slope of 2/3

Step-by-step explanation:

the rule is that you do y2-y1 over x2-x1 which you do 4-2 over 8-5 and your final answer you would get 2/3.

hope this helped!

i am new here and my name is arrianna call me arri for short

5 0
3 years ago
Read 2 more answers
A diameter of a circle has enpoints (-2, 10) and (6, -4) in the standard (x, y) coordinate plane. What is the center of the circ
wlad13 [49]

Answer:

The center of the circle is C(x,y) = (2, 3).

Step-by-step explanation:

The center of the circle is the midpoint of the segment between the endpoints. We can determine the location of the center by this vectorial expression:

C(x,y) = \frac{1}{2}\cdot R_{1}(x,y)+ \frac{1}{2}\cdot R_{2}(x,y) (1)

Where:

C(x,y) - Center.

R_{1} (x,y), R_{2} (x,y) - Location of the endpoints.

If we know that R_{1} (x,y) = (-2,10) and R_{2} (x,y) = (6,-4), then the location of the center of the circle is:

C(x,y) = \frac{1}{2}\cdot (-2,10)+\frac{1}{2}\cdot (6,-4)

C(x,y) = (-1, 5) + (3, -2)

C(x,y) = (2, 3)

The center of the circle is C(x,y) = (2, 3).

8 0
3 years ago
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