5.254071926×10^9
Standard form is taking a large number and putting it in a form where the number is smaller than 10. To do that (using the number above) you would take the first number which is 5 and then make every number after it part of a decimal. You then use [×10^x] (multiplied by 10 to the power of the amount of numbers after the decimal point (In this case 9)) to show others that to find the full number you multiply the decimal by 10 to the power of 9.
Answer:
500cm2
Step-by-step explanation:
2(9×20)+(7×20)
=360+140=500cm2
Answer: The answer is 50980
Step by Step Explanation:
First, temporarily assume that two letters with I are different, call them i 1 and i 2. Three "a" are also called as a 1, a 2 and a 3, and two h as h 1 and h 2. Then there are 11 * 10 * 9 * 8 * 7 = 55440 possible "words" (one of 11 is the first letter, 10 is the second, and so on). But because equal letters do the same "words," some "words" were counted twice or more. We have to deduct the number of "parasitic" counts although it is fairly small. The words that counted more than once are divided into many disjoint sets: 1) with two I but without repetitions of a and h; 2) with two h but without repetitions of a and I 3) with two a but without repetitions of I and h; 4) with three a but without repetitions of I and h; 5) with two I and two a's; 6) two i's and tree a's; 7) two i's and two h's 8) two h's and two a's; 9) two h's and one tree. The first category includes terms counted twice and its scale is (5 * 4) * (6 * 5 * 4) = 2400 (the first I stays at one of the 5 positions, the second at one of the 4, then 11-2i-1h-2a=6). So we have 2400/2 = 1200 to subtract. Group 2 gives -600 as well, and group 3 also. Group 4 gives * (6 * 5) = 1800 (5 * 4 * 3), and the terms are counted 6 times, -300. Groups 5, 7 , 8: 5 * 4 * 3 * 2 * 6 = 720 and counted four times, therefore -180. Group 6 and 9: 5 * 4 * 3 * 2 * 1 = 120, with 12 counts, -10. Altogether -(1200 * 3 + 300 + 180 * 3 + 10 * 2) = -4460.The answer will be 55440-4460 = 50980.
a. Recall the double angle identities:
Then
Applying the identity again, we have
as required.
b. Using the result from part (a),
(where 
is any integer)
Answer:
Step-by-step explanation:
