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olasank [31]
2 years ago
9

Solve for the perimeter of the figure below. Round your answer to the nearest hundredth. Show your work and explain the steps

Mathematics
1 answer:
vagabundo [1.1K]2 years ago
7 0

Answer:

18.87 units

Explanation:

the estimated distance from a to d is the same as a to b. we can find the length of the diagonal using pythagoras' theorem:

a² + b² = c²

√(2² + 3²) = 3.6056 (to 5s.f.)

now, we can use the same method to calculate the distance between d and c, as well as c and b:

√(3² + 5²) = 5.8310 (to 5s.f.)

adding these values, we will get:

(3.6056 + 5.8310) × 2 = <u>18.87 units</u> (to the nearest hundredth)

i hope this helps! :D

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Answer:

nuber 1

Simplifying

3x + 2y = 35

Solving

3x + 2y = 35

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-2y' to each side of the equation.

3x + 2y + -2y = 35 + -2y

Combine like terms: 2y + -2y = 0

3x + 0 = 35 + -2y

3x = 35 + -2y

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x = 11.66666667 + -0.6666666667y

Simplifying

x = 11.66666667 + -0.6666666667y

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Answer:

1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that \sqrt{3}=\frac{m}{n} where  m and n are non negative integers, and the fraction \frac{m}{n} is irreducible, i.e., the numbers m and n have no common factors.

Now, squaring the equality at the beginning we get that

3=\frac{m^2}{n^2} (1)

which is equivalent to 3n^2=m^2. From this we can deduce that 3 divides the number m^2, and necessarily 3 must divide m. Thus, m=3p, where p is a non negative integer.

Substituting m=3p into (1), we get

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Thus, 3 divides n^2 and necessarily 3 must divide n. Hence, n=3q where q is a non negative integer.

Notice that

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2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, r\in\mathbb{Q}, which is equivalent to say that r=\frac{m}{n} where  m and n are non negative integers. Also, assume that k\in\mathbb{Z}. So, we want to prove that k\cdot r\in\mathbb{Z}. Recall that an integer k can be written as

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Then,

k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}.

Notice that the product mk is an integer. Thus, the fraction \frac{mk}{n} is a rational number. Therefore, k\cdot r\in\mathbb{Q}.

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Write q=x+p and let us suppose that q is a rational number. So, we get that

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But the subtraction or addition of two rational numbers is rational too. Then, the number x must be rational too, which is a clear contradiction with our hypothesis. Therefore, x+p is irrational.

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B- 1.23 You can do that whit an app name photomath :D
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