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Artemon [7]
4 years ago
8

What is 7 over 9 subtract 1 over 3

Mathematics
2 answers:
Firlakuza [10]4 years ago
5 0
The answer is 4/9 or 0.44444
nevsk [136]4 years ago
3 0
So if i understand correctly this is fraction subtractions so in order to do it you have to have a common denominator and in order to do it you will change 1/3 to 3/9 and then subtract it from 7/9
7/9-1/3
7/9-3/9
4/3
i hope this helped 

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Answer:

cos x = 8/f.

Step-by-step explanation:

We can use the identity :

tan x = sin x / cos x.

Substituting:

e/8 = e/f / cos x

e/8 * cos x = e/f

cos x = e/f * 8/e

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8 0
3 years ago
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4 years ago
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Vladimir79 [104]

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7 0
3 years ago
(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?
DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500

  • There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111

  • There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611

7 0
3 years ago
What number should be added to 51 to get a sum -21
Anastaziya [24]
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3 years ago
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