Hey!
Finding the factorial for the first few numbers, we have:
1!=1
2!=2
3!=6
4!=24
5!=120
6!=720
7!=720*7
8!=720*56
What we can see as a clear pattern from 5! and on is that our number ends with a 0, making the units digit 0. Therefore, when we add the units digit of 5! and on, we have a result of 0. So, we can simply add the units digits of 1!, 2!, 3!, and 4!, which is 1+2+6+4=13. Since the units digit is the last number, we can drop the tens place to get an answer of 3.
Feel free to ask further questions!
To solve this problem you must apply the proccedure shown below:
1. You have the following logarithm:
<span>log(2)n=4
2. Therefore, you con rewrite it as below:
loga(b)=logb/loa
</span>
3. Therefore, you have:
log(2)n=4⇒log(n)/log(2)=4
4. Then, you obtain:
log(n)=4log(2)
5. Therefore, as you can see, the answer for the exercise shown above is the last option, which is:
log(n)=4log(2)
We can't tell, because you won't let us SEE
the electric meter shown in the exam figure.
Answer:
18
Step-by-step explanation:
Product of 7 and a number is 126.
Let's form the equation,
→ 7 × x = 126
The required value of x will be,
→ 7 × x = 126
→ x = 126/7
→ [ x = 18 ]
Hence, the number is 18.
