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Dafna11 [192]
3 years ago
7

Find the sides of a rectangle having area 125 square units if the sides are in the ratio 5:3.

Mathematics
1 answer:
tatiyna3 years ago
7 0

For this case we have the following data:

Area of the rectangle, A = 125

Relationship of the sides 5: 3

By definition, the area of a rectangle is given by:

A = a * b

Where

  • a: is the longest side.
  • b: it is the side of least length.

If the sides have a 5: 3 ratio, it means:

\frac{a}{b}=\frac{5}{3}

a=\frac{5}{3}b

Substituting in the formula of the area, we have:

125 =a * b\\

125=\frac{5}{3}b*b

Clearing b:

3 * 125 = 5 * b ^ 2

\frac{375}{5}= b ^ 2

75 = b ^ 2

b =\sqrt{75}b = 8.7 units

Substituting and clearing:

125 = a * 8.7

a =\frac{125}{8.7}

a = 14.4units

So, the sides are:

a = 14.4units

b = 8.7units

Answer:

a = 14.4units

b = 8.7units


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A trinomial is a perfect square if the square root of the first term times the square root of the third term times 2 equals the middle term.

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a) Adding 1/9:

\cdot \: \mathsf{x^{2} + \dfrac{2}{3} x + \dfrac{1}{9}} \\ \\ \\ \mathsf{\sqrt{x^{2}} \times \sqrt{\dfrac{1}{9}} \times 2 =} \\ \\ \\ \mathsf{x \times \dfrac{1}{3} \times 2 =} \\ \\ \\ \mathsf{\dfrac{2}{3} x \rightarrow it \: is \: a \: perfect \: square \: trinomial}

b) Adding 4/9:

\cdot \: \mathsf{x^{2} + \dfrac{2}{3} x + \dfrac{4}{9}} \\ \\ \\ \mathsf{\sqrt{x^{2}} \times \sqrt{\dfrac{4}{9}} \times 2 =} \\ \\ \\ \mathsf{x \times \dfrac{2}{3} \times 2 =} \\ \\ \\ \mathsf{\dfrac{4}{3} x \rightarrow it \: is \: not \: a \: perfect \: square \: trinomial}

c) Adding 4 and 1/9:

\cdot \: \mathsf{x^{2} + \dfrac{2}{3} x + \dfrac{1}{9} + 4 = x^{2} + \dfrac{2}{3} x + \dfrac{37}{9}} \\ \\ \\ \mathsf{\sqrt{x^{2}} \times \sqrt{\dfrac{37}{9}} \times 2 =} \\ \\ \\ \mathsf{x \times \dfrac{\sqrt{37}}{3} \times 2 =} \\ \\ \\ \mathsf{\dfrac{2\sqrt{37}}{3} x \rightarrow it \: is \: not \: a \: perfect \: square \: trinomial}

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Answer:

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Step-by-step explanation:

We can use the following kinematic equations:

\displaystyle v_f  = v_0 + at \text{ and } \Delta d = v_0 t + \frac{1}{2} at^2

We want to determine the distance the object has dropped after falling from rest and reaching an instantenous speed of 9 m/s.

We are given that the acceleration due to gravity is 9.8 m/s².

Using this fact and the first equation, find the time for which it took the object to reach 9 m/s. Note that the initial velocity is 0 m/s since the object started from rest.

\displaystyle \begin{aligned} v_f = v_0 + at \\ \\ (9\text{ m/s}) & = (0\text{ m/s}) + (9.8\text{ m/s$^2$})t \\ \\ 9\text{ m/s} & = (9.8 \text{ m/s$^2$})t \\ \\ t &\approx0.9184 \text{ s} \end{aligned}

To find how far the object dropped, we can use the second equation:

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In conclusion, the object would have dropped about 4.1323 meters.

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