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Dovator [93]
2 years ago
8

Which of the following fractions are equivalent to 15/21? Select all that apply.

Mathematics
2 answers:
nikklg [1K]2 years ago
4 0

Answer:

A and B

Step-by-step explanation:

  • There is an infinity number of equivalent fractions to 15/21.

To find an equivalent fraction to 15/21 , or to any other fraction, you just need to multiply (or divide, if the fraction is not yet reduced), both the numerator and the denominator of the given fraction by any non-zero natural number. For example:

  • By dividing the original fraction by 3, we get:

\frac{15/3}{21/3} =\frac{5}{7}

  • By multiplying the original fraction by 2, we get:

\frac{15*2}{21*2} =\frac{30}{42}

  • Here is a list of the equivalent fractions of 15/21

5/7, 10/14, 15/21, 20/28, 25/35, 30/42, 35/49, 40/56, 45/63, 50/70, 55/77, 60/84, 65/91, 70/98, 75/105, 80/112, 85/119, 90/126, 95/133, 100/140 ...

Answer:

Therefore, the answer is <u>A</u> and <u>B</u>.

scoundrel [369]2 years ago
3 0

Answer:

A. 5/7
B. 30/42

Step-by-step explanation:

I just used an equivalent fractions calculator.
Please correct me if I'm wrong :)

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Among all right triangles whose hypotenuse has a length of 12 cm, what is the largest possible perimeter?
Veronika [31]

Answer:

Largest perimeter of the triangle =  

P(6\sqrt{2}) = 6\sqrt{2} + \sqrt{144-72} + 12 = 12\sqrt{2} + 12 = 12(\sqrt2 + 1)

Step-by-step explanation:

We are given the following information in the question:

Right triangles whose hypotenuse has a length of 12 cm.

Let x and y be the other two sides of the triangle.

Then, by Pythagoras theorem:

x^2 + y^2 = (12)^2 = 144\\y^2 = 144-x^2\\y = \sqrt{144-x^2}

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P(x) = x + \sqrt{144-x^2} + 12

where P(x) is a function of the perimeter of the triangle.

First, we differentiate P(x) with respect to x, to get,

\frac{d(P(x))}{dx} = \frac{d(x + \sqrt{144-x^2} + 12)}{dx} = 1-\displaystyle\frac{x}{\sqrt{144-x^2}}

Equating the first derivative to zero, we get,

\frac{dP(x))}{dx} = 0\\\\1-\displaystyle\frac{x}{\sqrt{144-x^2}} = 0

Solving, we get,

1-\displaystyle\frac{x}{\sqrt{144-x^2}} = 0\\\\x = \sqrt{144-x^2}}\\\\x^2 = 144-x^2\\\\x = \sqrt{72} = 6\sqrt{2}

Again differentiation P(x), with respect to x, using the quotient rule of differentiation.

\frac{d^2(P(x))}{dx^2} = \displaystyle\frac{-(144-x^2)^{\frac{3}{2}}-x^2}{(144-x)^{\frac{3}{2}}}

At x = 6\sqrt{2},

\frac{d^2(V(x))}{dx^2} < 0

Then, by double derivative test, the maxima occurs at x = 6\sqrt{2}

Thus, maxima occurs at x = 6\sqrt{2} for P(x).

Thus, largest perimeter of the triangle =  

P(6\sqrt{2}) = 6\sqrt{2} + \sqrt{144-72} + 12 = 12\sqrt{2} + 12 = 12(\sqrt2 + 1)

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