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olchik [2.2K]
3 years ago
11

(Economics) Select all that apply.

Mathematics
2 answers:
lesya [120]3 years ago
7 0

A. final goods and services

D. new products

m_a_m_a [10]3 years ago
6 0
<span>the right answer is a.) final goods and sevices

</span>
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What is the value of s?<br><br> ? units
Anton [14]
Pythagoras theorem:
hypotenuse²=leg₁²+leg₂²

Data:
leg₁=8
leg₂=15
hypotenuse=s

hypotenuse²=(8)²+(15)²
hypotenuse²=64+225
hypotenuse²=289
hypotenuse=√289
hypotenuse=17

Answer: s=17
8 0
3 years ago
Read 2 more answers
Solve the system 2x+y=3 and -2y=14-6 write each equation in slope form
frozen [14]

Answer:

First - y=-2x+3

Second - y=-7+3

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Help pleaseeee! ASAP
34kurt

Answer:

c

Step-by-step explanation:

5 0
3 years ago
There is no smallest positive rational number because, if there were, then it could be divided by two to get a smaller one. EXPL
Gennadij [26K]

Answer:

Since a/2⁽ⁿ ⁺ ¹⁾b <  a/2ⁿb,  we cannot find a smallest positive rational number because there would always be a number smaller than that number if it were divided by half.

Step-by-step explanation:

Let a/b be the rational number in its simplest form. If we divide a/b by 2, we get another rational number a/2b. a/2b < a/b. If we divide a/2b we have a/2b ÷ 2 = a/4b = a/2²b. So, for a given rational number a/b divided by 2, n times, we have our new number c = a/2ⁿb where n ≥ 1

Since \lim_{n \to \infty} \frac{a}{2^{n}b } = a/(2^∞)b = a/b × 1/∞ = a/b × 0 = 0, the sequence converges.

Now for each successive division by 2, a/2⁽ⁿ ⁺ ¹⁾b <  a/2ⁿb and

a/2⁽ⁿ ⁺ ¹⁾b/a/2ⁿb = 1/2, so the next number is always half the previous number.

So, we cannot find a smallest positive rational number because there would always be a number smaller than that number if it were divided by half.

3 0
3 years ago
Find the permeter of a quadrilateral with vertices at C (2,4), D(1,1), E(5,0), and F(3,4). Round your answer to the nearest hund
denpristay [2]

Answer:

12.76

Step-by-step explanation:

This is much easier to solve if you draw it out on a graph. The perimeter is the sum of all the sides. I used the Pythagorean theorem to find the hypotenuse.

{a}^{2}  +  {b}^{2}  =  {c}^{2}  \\ c =  \sqrt{ {a}^{2} +  {b}^{2}  }

dc =  \sqrt{ {3}^{2}  +  {1}^{2} }  =  \sqrt{10}  \\  cf = 1 \\ fe =  \sqrt{ {4}^{2}  +  {2}^{2} }  =  \sqrt{20}  \\ de =  \sqrt{ {1}^{2} +  {4}^{2}  }  =  \sqrt{17}  \\  \sqrt{10}  + 1 +  \sqrt{20}  +  \sqrt{17}  = 12.76

7 0
3 years ago
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