Here's a simple way to solve it: 2*3=6 6*3=18 18*3=54 54*3=162 162*3=486 486*5=2,430 2,430*7=17,010 In simpler terms, 2*3*3*3*3*3*5*7=17,010.
Any integer of which both 5 and 7 are factors.
To find the number of different 7-letter permutations that can be formed from 5 identical H's and 2 identical T's, we use the formula for permutations of multiset: [ \frac{n!}{n_1! \times n_2!} ] where (n) is the total number of letters, and (n_1) and (n_2) are the counts of each type of letter. Here, (n = 7), (n_1 = 5) (for H's), and (n_2 = 2) (for T's). Thus, the calculation is: [ \frac{7!}{5! \times 2!} = \frac{5040}{120 \times 2} = \frac{5040}{240} = 21 ] Therefore, there are 21 different permutations.
The event occurs 6 times daily, 4 times weekly, 2 times monthly, and 1 time yearly.
There are 2 inches between the 5 and 7 marks on a ruler.
14
2 * 2 * 5 * 7 * 11 = 1540
2*2 = 4 * 3 = 12 * 5 = 60 * 5 = 300 * 7 = 2,100
To express (2^{-5} \times 28) as an exponential expression, we can first rewrite 28 in terms of base 2. Since (28 = 4 \times 7 = 2^2 \times 7), we can substitute this into the expression: [ 2^{-5} \times 28 = 2^{-5} \times (2^2 \times 7) = 2^{-5 + 2} \times 7 = 2^{-3} \times 7. ] Thus, the exponential expression is (2^{-3} \times 7).
To rewrite (2 \times 5 \times 5 \times 7) using exponents, you can express the repeated multiplication of the number 5 as (5^2). Therefore, the expression can be rewritten as (2 \times 5^2 \times 7).
2 * 3 * 3 * 5 * 7 = 630
2 x 2 x 5 x 7 x 11 = 1540
The expression (2 \times 2 \times 2 \times 5 \times 5 \times 7) can be written in exponent form as (2^3 \times 5^2 \times 7^1). This indicates that 2 is multiplied three times, 5 is multiplied two times, and 7 is multiplied once.
350
2 * 3 * 5 * 7 = 210
7*7*2*2*5*3 =(7*7)*(2*3)*(2*5) =49*6*10 =(50-1)*6*10 =(50*6-6)*10 =(300-6)*10 =294*10 =2940
1540 has the prime factorization of 2 times 2 times 5 times 7 times 11
2 x 3 x 5 x 7 = 210