What is the factorization of 6x2-19x-55
1 answer:
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We can factorise this expression by grouping. First let us arrange it in this way
Let us bracket them like this:
In the first bracket portion, take x as common and in the second expression, take 2 as common.
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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,
- 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,
- 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: