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Neko [114]
3 years ago
15

if a 25 ft rope is cut into two sections with one section four times as long as the other section, how long is the shorter piece

?
Mathematics
2 answers:
Ganezh [65]3 years ago
4 0

Total length of rope = 25 feet

Let us say one part of the rope is x, then the other part will be 25-x.

We are given that one section of the rope is four times as long as the other.

So let us make an equation using this information,

x=4(25-x)

x=100-4x

5x=100

x=20

So one part is 20 feet and the second part is 25-20= 5 feet.

Answer: The shorter part of the rope is 5 feet long.


grin007 [14]3 years ago
3 0
3.125 ft long or 3.13 ft long
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Answer:

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Step-by-step explanation:

Since 45% of the participants were female and there were a total of 80 students, 36 of the 80 students were female. That means that 44 of the students were male.

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It's 57 and 59
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4 years ago
The compound interest on a sum of money in
Yakvenalex [24]

Answer:

The difference between the principal and the compound interest in three years is Rs 17,994

Step-by-step explanation:

The compound interest is given according to the following formula;

C.I. = P \cdot \left ( 1 + \dfrac{r}{n} \right ) ^{n\cdot t} - P

The given amount of the compound after 2 years = Rs 5,460

The given amount of the compound after 4 years = Rs 12,066.60

Therefore, we have;

5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P...(1)

12,066.60 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{4} - P ...(2)

Dividing equation (2) by (1), we have;

\dfrac{12,066.60}{5,460} = \dfrac{P \cdot \left ( \left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1\right )}{P \cdot \left (\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1 \right ) } =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1  }

Let \  \left ( 1 + \dfrac{r}{100} \right ) ^{2} = x, we \ get;

\dfrac{12,066.60}{5,460} =\dfrac{\left ( 1 + \dfrac{r}{100} \right ) ^{4} - 1}{\left ( 1 + \dfrac{r}{100} \right ) ^{2} -1  } = \dfrac{x^2 - 1}{x - 1}

∴ 12,066.60 × (x - 1) = 5,460 × (x² - 1) = 5,460 × (x - 1) ×(x + 1)

∴ 12,066.60 × (x - 1)/(x - 1) = 5,460 × (x + 1)

12,066.60/5,460 = x + 1

x =  12,066.60/5,460 - 1 = 1.21 = 121/100

x = 121/100

\left ( 1 + \dfrac{r}{100} \right ) ^{2} = x = \dfrac{121}{100}

1 + \dfrac{r}{100}   =\sqrt{ \dfrac{121}{100}} = \dfrac{11}{10}

We get

\dfrac{12,066.60}{5,460} =\dfrac{221}{100}

\therefore \dfrac{12,066.60}{5,460} =\dfrac{221}{100} = \left ( 1 + \dfrac{r}{100} \right ) ^{2}

1 + \dfrac{r}{100} = \sqrt{ \dfrac{221}{100} } = \dfrac{\sqrt{221} }{10}

\dfrac{r}{100} = \dfrac{\sqrt{221} }{10} - 1

\dfrac{r}{100}   = \dfrac{11}{10} - 1 = \dfrac{1}{10} = 0.1

r = 100 × 0.1 = 10%

r = 10%

Therefore, we have;

5,460 = P \cdot \left ( 1 + \dfrac{r}{100} \right ) ^{2} - P = P \times \left ( 1 + 0.1\right ) ^{2} - P

5,460  = P \times \left ( 1 + 0.1\right ) ^{2} - P = P \times \left (\left ( 1 + 0.1\right ) ^{2} - 1\right) = P \times \dfrac{21}{100}

P = \dfrac{100}{21} \times 5,460 = 26,000

The principal = Rs. 26,000

The compound interest in 3 years is therefore;

CI_3 = 26,000 \times \left ( 1 + \dfrac{10}{100} \right ) ^{3} - 26,000= 8606

The difference, 'd', between the principal and the compound interest in three years, is given as follows;

d = P - CI₃

d = 26,600 - 8606 = 17994

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you need to follow these steps to be able to solve any equation

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Step-by-step explanation:

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