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Anuta_ua [19.1K]
2 years ago
6

9/20+3/18 i need help witth this question

Mathematics
2 answers:
Ugo [173]2 years ago
8 0

Answer:

111/180

Step-by-step explanation:

You find the lcm (loswest common denominator), which is 180. In order to get 180, you times 9 to 20, 10 to 18.

We have to times on the numerator and denominator.

add them together.

111/180

Hope this helps!

Nataliya [291]2 years ago
3 0

Answer:

37/60 OR 111/180

(37/60 is simplified and 111/180 is the original)

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Answer:

B

Step-by-step explanation:

4 0
3 years ago
Solve for x in the equation x2 - 12x+ 36 = 90
12345 [234]

Answer:

6 + 3\sqrt{10\\} or 6 - 3\sqrt{10\\}

Step-by-step explanation:

x^2 - 12x + 36 = 90

x^2 - 12x - 54 = 0

x = \frac{-b +/- \sqrt{b^{2} + 4ac }}{2a}

a = 1, b= -12, c= -54

6 + 3\sqrt{10\\} or 6 - 3\sqrt{10\\}

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3 years ago
Suppose each edge of a cube is 7.8 cm long. Which is the BEST estimate for the volume of the cube?
Naya [18.7K]
Multiply like this. And it gives you your answer.





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6 0
2 years ago
Read 2 more answers
(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

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3 years ago
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Answer:

27.69

Step-by-step explanation:

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3 years ago
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