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k0ka [10]
2 years ago
14

A number whose value is greater than zero

Mathematics
2 answers:
weeeeeb [17]2 years ago
8 0

Natural numbers because natural numbers are 1-infinity

Dima020 [189]2 years ago
6 0

Answer:

natural number

Step-by-step explanation:

natural number

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Choose the solution set. Given: x - 8 > -3.
charle [14.2K]

Answer:

x> 5

Step-by-step explanation:

x - 8 > -3

Add 8 to each side

x - 8+8 > -3+8

x> 5

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3 years ago
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Drag the numbers below to put them in order from least to greatest: 4 16 1 7 -2 -20​
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Are the 1 and 7 rogether? because if so then -20, -2, 4, 16, 17, if not then -20, -2, 1, 4, 7, 16

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Step-by-step explanation:

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Anyone knows how to do questions 7 and 8? 15 pts!!
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7) Certainly there is a typo in the statement, just see that the expression of item (ii) is different from that of item (i). Probably the correct expression is: 2x^2-4x+5. With this consideration, we can continue.

(i) Let E the expression that we are analyzing:

E=2x^2-4x+5\\\\ E=2x^2-4x+2-2+5\\\\ E=2(x^2-2x+1)-2+5\\\\ E=2(x-1)^2+3

Since (x-1)² is a perfect square, it is a positive number. So, E is a result of a sum of two positive numbers, 2(x-1)² and 3. Hence, E is a positive number, too.

(ii) Manipulating the expression:

2x^2+5=4x\\\\ 2x^2-4x+5=0

So, it's the case when E=0. However, E is always a positive number. Then, there is no real number x that satisfies the expression.

8) Let E the expression that we want to calculate:

E=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)+1\\\\ E-1=(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)

Multiplying by (2-1) in the both sides:

(2-1)(E-1)=(2-1)(2+1)(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2-1)(2+1)}_{2^2-1}(2^2+1)(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ (E-1)=\underbrace{(2^2-1)(2^2+1)}_{2^4-1}(2^4+1)\cdot ...\cdot(2^{32}+1)\\\\ ... Repeating the process, we obtain: ...\\\\ E-1=(2^{32}-1)(2^{32}+1)\\\\ E-1=2^{64}-1\\\\ \boxed{E=2^{64}}
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3 years ago
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