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Archy [21]
3 years ago
6

Type the base. 1/(7 ^ 2)​

Mathematics
1 answer:
Butoxors [25]3 years ago
7 0

Answer:

,,,,,,,,,,,,,, mmmmm mm m m m

Step-by-step explanation:

 m mm m m m mm m

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\sqrt{12}  -  \sqrt{48}  =

\sqrt{12}  -  \sqrt{4 \times 12}  =

\sqrt{4 \times 3}  -  \sqrt{4 \times 4 \times 3}  =

\sqrt{ {2}^{2} \times 3 }  -  \sqrt{ {4}^{2} \times 3 }  =

2 \sqrt{3}  -  4\sqrt{3}  =

(2 - 4) \sqrt{3}  =

- 2 \sqrt{3}

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8 0
3 years ago
How many whole numbers between 99 and 999 contain the digit '1'?
Sedbober [7]

Answer:

162

Step-by-step explanation:

The number with "1" digit starts with 101 then 110 and so on.

6 0
3 years ago
Read 2 more answers
Please give me the answer nothing else
Nookie1986 [14]

Answer:

12.5

Step-by-step explanation:

12x6 is 12 + 5 meters = 12.5m

3 0
2 years ago
What’s the correct answer for this?
otez555 [7]

Answer:

0.7 + 0.4 - 0.2 = 0.9

Step-by-step explanation:

Let's denote the probabilities as following:

The probability that the show had animals is

P(A) = 0.7

The probability that the show aired more than 10 times is

P(B) = 0.4

The probability that the show had animals and aired more than 10 times is

P(A⋂B) = 0.2

The probability that a randomly selected show had animals or aired more than 10 times is P(A⋃B)

The correct form of  addition rule to determine the probability that a randomly selected show had animals or aired more than 10 times is:

P(A⋃B) = P(A) + P(B) - P(A⋂B) = 0.7 + 0.4 - 0.2 = 0.9

=> Option B is correct

Hope this helps!

7 0
3 years ago
About Exercise 2.3.1: Proving conditional statements by contrapositive Prove each statement by contrapositive
Sladkaya [172]

Answer:

See proofs below

Step-by-step explanation:

A proof by counterpositive consists on assuming the negation of the conclusion and proving the negation of the hypothesis.

a) Assume that n is not odd. Then n is even, that is, n=2k for some integer k. Hence n²=4k²=2(2k²)=2t for some integer t=2k². Then n² is even, therefore n² is not odd. We have proved the counterpositive of this statement.

b) Assume that n is not even, then n is odd. Thus, n=2k+1 for some integer k. Now, n³=(2k+1)³=8k³+6k²+6k+1=2(4k³+3k²+3k)+1=2t+1 for the integer t=4k³+3k²+3k. Thus n³ is odd, that is, n³ is not even.

c) Suppose that n is not odd, that is, n is even. Now, n=2k for some integer k. Then 5n+3=10k+3=2(5k+1)+1, thus 5n+3 is odd, then 5n+3 is not even.

d) Suppose that n is not odd, then n is even. Now, n=2k for some integer k. Then n²-2n+7=4k²-4k+7=2(2k²-2k+3)+1. Hence n²-2n+7 is odd, that is, n²-2n+7 is not even.

e) Assume that -r is not irrational, then -r is rational. Since -1 is rational, then (-1)(-r)=r is rational. Thus r is not irrational.

f) Assume that 1/z is not irrational. Then 1/z is rational. Multiplucative inverses of rational numbers are rational, hence z is rational, that is, z is not irrational.

g) Suppose that z>y. We will prove that z³+zy²≤z²y+y³ is false, that is, we will prove that z³+zy²>z²y+y³. Multiply by the nonnegative number z² in the inequality z>y to get z³>z²y (here we assume z and y nonzero, in this case either z³>0=y³ is true or z³=0>y³ is true). On the other hand, multiply by z² (positive number) to get zy²>y³. Add both inequalities to obtain z³+zy²>z²y+y³ as required.

h) Suppose than n is even. Then n=2k, and n²=4k² is divisible by 4.

i) Assume that "z is irrational or y is irrational" is false. Then z is rational and y is rational. Rational numbers are closed under sum, then z+y is rational, that is, z+y is not irrational.

3 0
3 years ago
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