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cricket20 [7]
4 years ago
7

Which relationship describes angles 1 and 2?

Mathematics
1 answer:
jenyasd209 [6]4 years ago
3 0
The only one that applies is supplementary angles
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Simplify the expression 8xy + 6.3h + 7xy - xy - 2h
solong [7]
Answer:

14xy+4.3h


Step-by-step:

= 8xy + 6.3h + 7xy - xy + -2h

= (8xy + 7xy + -xy) + (6.3h + -2h)

=14xy + 4.3h
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3 years ago
Factor the polynomial
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   3x⁶y² + 57x⁵y³ + 252x⁴y⁴
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= 3x⁴y²(x(x + 12y) + 7y(x + 12y))
= 3x⁴y²(x + 12y)(x + 7y)
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4 years ago
A college chemistry teacher wants to know whether taking a practice
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brainly.com/question/17278316
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3 years ago
Use the inverse of this function to select the correct graph.
xxTIMURxx [149]

Answer:

D

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x __ y

2  __2

3  __ 2.387

7__ 3

3 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
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