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Hitman42 [59]
3 years ago
9

What's the answer to this? and please explain.

Mathematics
2 answers:
vichka [17]3 years ago
5 0
<h3>Answer: No, there isn't enough ribbon</h3>

===========================================================

Use this formula

a & b/c = (a*c+b)/c

to convert from a mixed number to an improper fraction

So,

a & b/c = (a*c+b)/c

3 & 3/4 = (3*4+3)/4

3 & 3/4 = 15/4

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

Next, we add the fractions 15/4 and 5/6. The denominators aren't the same, so we can't add right away. We need to get the denominators to the LCD.

In this case, the LCD is 4*6 = 24

  • The fraction 15/4 becomes 90/24 after multiplying top and bottom by 6.
  • The fraction 5/6 becomes 20/24 after multiplying top and bottom by 4

Now we can add the fractions:

90/24+20/24 = (90+20)/24 = 110/24

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

So far, we've added the 3&3/4 portion to the 5/6 portion to get the result 110/24

The last thing to do is to add on the two 1/3 ft pieces, which means we're adding on 2/3 of a foot

2/3 = 16/24 after multiplying top and bottom by 8

Add this to the result of the last section

110/24+16/24 = (110+16)/24 = 126/24

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

From here, we convert the improper fraction 126/24 to decimal form

Using a calculator, you should find that 126/24 = 5.25

The result is larger than 5, which means she used too much ribbon. There won't be enough ribbon to do everything she wants to do.

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

The shortcut we can take is to type the following into the calculator

3+3/4+5/6+2/3

doing so should lead to 5.25

sasho [114]3 years ago
5 0

Answer:

Yes

Step-by-step explanation:

So at the begining, Cherly has 60 in (12inx5ft) of ribbon. She uses 45 in(3.75x12) to make the hair bow, leaving 15 in of ribbon. She, then, uses 10 in (5/6x12) of ribbon, leaving 5 in. After that, Cheryl uses 8 in [2(1/3x12)] to put on the picture frame, leaving two inches of ribbon.

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For a boat to float in a tidal bay, the water must be at least 2.5 meters deep. The depth of water around the boat, ????(????),
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a. Period, T = 12.57 hours. b. Latest time = 294.16 hours, 6 a.m 12 days later

Step-by-step explanation:

For a boat to float in a tidal bay, the water must be at least 2.5 meters deep. The depth of water around the boat, d(t), in meters, where t is measured in hours since midnight, is d(t) = 5 + 4.6sin(0.5t). (a) What is the period of the tides in hours? (b) If the boat leaves the bay at midday, what is the latest time it can return before the water becomes too shallow?

a. The period, T of the tides is gotten from ω = 2π/T, where ω is the angular frequency of the wave and T its period. Comparing the sine part of equation of the tides with the general equqtion of a sine wave, Asinωt, thus Asinωt = 4.6sin(0.5t),

so ω = 0.5 rad/s.

Equating this to 2π/T implies

0.5 = 2π/T

therefore, T = 2π/0.5 = 12.566 hours ≈ 12.57 hours

b. The tide becomes too low if it is less than 2.5 m i.e d(t) < 2.5 m. So, equating d(t) as 2.5 in the tidal equation, we have d(t) = 5 + 4.6sin(0.5t).

2.5< 5+ 4.6sin(0.5t)

2.5-5 < 4.6sin(0.5t)

-2.5 < 4.6sin(0.5t)

-2.5/4.6 < sin(0.5t)

-0.543 < sin(0.5t)

sin⁻¹(-0.543)<0.5t

-32.92<0.5t. Since we cannot have negative time, we add 180 to -32.92 to give 147.08

147.08<0.5t

t>147.08/0.5=294.16 hours

So, if t is greater than 294.16 hours, the water will be shallow. That is, below 2.5 m. which is 12 days 6hours 8 min 38 s which is 6 a.m 12 days later

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