Log3(7) + log3(x)
Terms that are multiplied inside one log become added when you separate them where each term gets its own log.
Long division: (x³ + 2) ÷ (x + 1)
<u> </u><u>x² – x + 1 </u>
x³ + 0x² + 0x + 2 | x + 1
<u>– x³ – x²</u> ⋮ ⋮
– x² + 0x ⋮
<u>+ x² + x</u><span> ⋮</span>
+ x + 2
<span> </span> <u>– x – 1</u>
+ 1
Quotient: Q(x) = x² – x – 1;
Remainder: R(x) = + 1.
I hope this helps. =)
Step-by-step explanation:
The statement in the above question is True.
Sum of three prime numbers (other than two) is always odd.
Going by Christian Goldbach number theory ,
- Goldbach stated that every odd whole number greater than 5 can be written as sum of three prime numbers .
Lets take an example,
- 3 + 3 + 5 = 11
- 3 + 5 + 5 = 13
- 5 + 5 + 7 = 17
Later on in 2013 the Mathematician <u>Harald Helfgott</u> proved this theory true for all odd numbers greater than five.
Answer:
The answer is 5G
Step-by-step explanation:
Answer:
Step-by-step explanation:
<u>Given:</u>
- AB = 192 cm
- AC : CB = 1 : 3
- CD = BC/12
- The distance between midpoints of AD and CB = x
<u>Find the length of AC and CB:</u>
- AC + CB = AB
- AC + 3AC = 192
- 4AC = 192
- AC = 192/4
- AC = 48 cm
<u>Find CB:</u>
<u>Find the length of CD:</u>
- CD = BC/12 = 144/12 = 12 cm
<u>Find the length of AD:</u>
- AD = AC - CD = 48 - 12 = 36 cm
<u>Find the midpoint of AD:</u>
<u>Find the midpoint of CB:</u>
- m(CB) = AC + 1/2CB = 48 + 144/2 = 48 + 82 = 130 cm
<u>Find the distance between the midpoints:</u>
