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creativ13 [48]
3 years ago
14

A l-meter steel rod is bent into a rectangle so that its length exceeds quadruple its width

Mathematics
1 answer:
Dmitry_Shevchenko [17]3 years ago
6 0

Answer:

The dimensions of the newly formed rectangle is length is 41 cm and width is 9 cm.

Step-by-step explanation:

Given,

A 1 meter steel rod is bent into a rectangle.

That means the perimeter of the rectangle is 1 meter.

So, Perimeter = 1\ m=100\ cm

Let the width of the rectangle be 'x'.

Now According to question, length exceeds quadruple its width by 5 cm.

Hence framing the above sentence in equation form, we get;

Length=4x+5

Now we use the formula of  perimeter of rectangle,

Perimeter=2(Length+width)

On substituting the values, we get;

2(4x+5+x)=100\\\\(5x+5)=\frac{100}{2}\\\\5x+5=50\\\\5x=50-5\\\\5x=45\\\\x=\frac{45}{5}=9

Now, we substitute the value of x to get the value of length;

Length=4x+5=4\times9+5=36+5=41\ cm

Thus, Width=9 cm    Length=41 cm

Hence the dimensions of the newly formed rectangle is length is 41 cm and width is 9 cm.

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podryga [215]

Complete question is;

There are six professors teaching the introductory discrete mathematics class at a university. The same final exam is given by all six professors. If the lowest possible score on the final is 0 and the highest possible score is 100.

How many students must there be to guarantee that there are two students with the same professor who earned the same final examination score?

Answer:

607 students

Step-by-step explanation:

To solve this question means we have to use Pigenhole principle which states that if y number of items are put into z number of containers, with y > z, then it means that that at least one container must definitely contain more than one item.

Now, let's find the number of boxes and objects:

If possible scores are from 0 to 100 with both inclusive it means number of possible scores = 101.

Now, If there was only one professor grading the students, in order to ensure that that there are two students with the same professor who earned the same final examination score,

The number of students would have to be = 101 + 1 = 102 students

Meanwhile, for each student, since there are 6 professors, the possible combination for a score will be = 6 possible combinations.

Therefore, number of students that will guarantee that there are two students with the same professor who earned the same final examination score must be a minimum of: (6 × 101) + 1 = 607

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