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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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Q: How many two digit numbers are divisible by 7?
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Related questions

How many two - digit numbers are divisible by 3?

There are 30 such numbers.


How many two digit numbers are divisible by either 3 or 5?

How many two digit number are divisible by 5


How many two digit numbers are divisible by 3?

30 of them.


How many 2 digit numbers are divisible by 3?

Starting at 12 and ending at 99, there are 30 two-digit numbers divisible by three.


What is the probability that a four digit number is divisible by 2?

Even numbers are divisible by two, and half of all numbers are even, so there is a 50 percent chance that a four-digit number is divisible by two.


How many two digit numbers divisible by neither 3 nor 5?

48.The first two digit number is 10, the last two digit number is 99, so there are 99 - 10 + 1 = 90 two digit numbers→ 10 ÷ 3 = 31/3 → first two digit number divisible by 3 is 4 x 3 = 12→ 99 ÷ 3 = 33 → last two digit number divisible by 3 is 33 x 3 = 99→ 33 - 4 + 1 = 30 two digit numbers divisible by 3→ 10 ÷ 5 = 2 → first two digit number divisible by 5 is 2 x 5 = 10→ 99 ÷ 5 = 194/5 → last two digit number divisible by 5 is 19 x 5 = 95→ 19 - 2 + 1 = 18 two digit numbers divisible by 5→ 30 + 18 = 48 two digit numbers divisible by 3 or 5 OR BOTH.The numbers divisible by both are multiples of their lowest common multiple: lcm(3, 5) = 15, and have been counted twice, so need to be subtracted from the total→ 10 ÷ 15 = 010/15 → first two digit number divisible by 15 is 1 x 15 = 15→ 99 ÷ 15 = 69/15 → last two digit number divisible by 15 is 6 x 15 = 90→ 6 - 1 + 1 = 6 two digit numbers divisible by 15 (the lcm of 3 and 5)→ 48 - 6 = 42 two digit numbers divisible by 3 or 5.→ 90 - 42 = 48 two digit numbers divisible by neither 3 nor 5.


What is the sum of all two digit numbers which are divisible by 5?

945


How many 2 digit numbers are divisible by 5 and 6?

If n is divisible by both 5 and 6, then it should be divisible by 30 (5 * 6). Considering you are asking for only two-digit numbers, the answer(s) would be 30, 60, and 90. So, three numbers.


How many two-digit numbers are divisible by both 9 and 2?

There are five numbers: 18, 36, 54, 72, 90.


How many 2-digit or 3-digit numbers can be formed using the digits 134568 and 9 which are diisible by 4?

48. Assuming no digit can be used more than once, the two digit numbers divisible by 4 are: 16, 36, 48, 56, 64, 68, 84, 96 8 of them. For any number to be divisible by 4, only the last two digits need be divisible by 4; so for three digit numbers, each of the two digit numbers above can be preceded by any of the remaining 5 digits and still be divisible by 4. → 5 x 8 = 40 three digit numbers are divisible by 4 → 40 + 8 = 48 two or three digit numbers made up of the digits {1, 3, 4, 5, 6, 8, 9} are divisible by 4. If repeats are allowed, there are an extra 2 two digit numbers (44 and 88) and each of the two digit numbers can be preceded by any of the 7 digits, making a total of 7 x 10 + 10 = 80 two and three digits numbers divisible by 4 make up of digits from the given set.


How many four digit numbers which are divisible by 7?

Best way to work this out: find the highest number below 10,000 that is divisible by 7 (9,996) and divide that by 7 (1,428). 1,428 is the amount of one-, two-, three- and four-digit numbers divisible by 7. Now find the number of one-, two- and three-digit numbers divisible by 7 (which is 994/7 = 142) and subtract this number from 1,428, giving us 1,286.


What two digit numbers greater than 74 are divisible by 2 and 3?

78,84,90,96

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