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san4es73 [151]
3 years ago
6

16. You want to decorate around the top of a jar with ribbon. The length around the jar is

Mathematics
1 answer:
VladimirAG [237]3 years ago
3 0

Answer: A) We need 1 1/2 ft of ribbon

B)$1.60 per foot

C)$2.40 is the total

Step-by-step explanation:

So if the length around the jar is 18 inches and we know that there is 12 inches in a foot, and 3 feet in a yard and $ 4.80 per every 3 feet. These become basis numbers. 18in- 12inch= 6 inch. So now we have 1 1/2 feet of ribbon to cover the circumference of the jar. Now we are left to answer the second question how much does it cost per foot? We can divide 4.80 by 3 to see the total cost of a foot of ribbon. That turns to 1.60 per foot. We divide that by two to cover the 6 inches. And the answer is $2.40 for 1 1/2 feet of ribbon

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Can anybody help out this math question it’s about graph please
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Neither one of the slopes are going to be minus. It means you are travelling backwards in time, which is wonderful if you are a sci-fi fan, but not so good if you are Sharon. A and D has Sharon going from 70 to 0. That can't be happening so both are wrong.


Now you have to decide between B and C. The intersection point has Sharon going upwards until she is 20. She started out at 70. The graph has John starting at 70. That's not right.


So we've eliminated A,D and now C.


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2 years ago
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inna [77]

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Step-by-step explanation:

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7 0
3 years ago
Read 2 more answers
Given f(x)=cos c and g(x)=cot x, what are the domain and range of f(g(x))?
ololo11 [35]

Answer:

Alternative C is the correct answer

Step-by-step explanation:

The first step is to determine the composite function;

f[g(x)]

f[g(x)]=cos[cot(x)]

We then employ a graphing utility to determine the range and the domain of the new function.

The range is the set of y-values for which the function is defined. In this case it is;

[-1,1]

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8 0
2 years ago
At noon, the minute and the hour hands of a clock overlap. In how long will they again overlap?
kati45 [8]

Answer:

Okay. At noon, the hour and minute hand overlap. First, let’s calculate the speed of the hands.

The minute hand rotates once around the clock every hour. It goes 360 degrees in 60 minutes, so it rotates 360/60 or 6 degrees per minute. The hour hand goes once around the clock every 12 hours. Because there are 60 minutes in an hour, it goes 360 degrees 12*60 or 720 minutes. This means it travels at a speed of 360/720 or 0.5 degrees per minute.

Once the clock starts, the minute hand goes ahead and rotates ahead of the hour hand. At 1:00, the minute hand is at 12, and the hour hand is ahead at 1. This is where we will start our calculations.

Now, we have a variable, m, which is the minutes that have passed since the clock started. So, our minute hand speed is 6m, and our hour hand speed is 0.5m. However, because the hour hand is at one, it is at 30 degrees. (360/12 * 1.) This means our equation is 6m = 0.5m+30.

Now, we calculate.

6m = 0.5m+30

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So at 5.45 minutes, the minute hand is at the hour hand. However, because one hour has passed since we started, we have to add on one hour.

So, our final total is 1 hour 5 minutes and 27 seconds.

Step-by-step explanation:

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3 years ago
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