By induction:
It's true for
, since
clearly contains a factor of 3.
Suppose it's true for
, that
is divisible by 3. Then
where
is an integer. This reduces to
and both terms are clearly multiples of 3. We know that
is an integer since we had set
previously, which implies
is a multiple of
. So the statement is true.
9514 1404 393
Answer:
(a) f(3) < g(3)
Step-by-step explanation:
Put the number where the variable is and do the arithmetic.
f(3) = 4·3³ = 4·27 = 108
g(3) = 3·4³ = 3·64 = 192
108 < 192
f(3) < g(3)
7/9 of 3/14 is 1/6.
Tell me if I'm wrong.
Hope this helps :)
Answer:
Step-by-step explanation:
11) First write in decreasing exponential terms and fill in blanks with zeros.
Our goal is to eliminate all term in the dividend by subtraction
________________
2v - 2 | 2v³ - 16v² + 0v + 13
we see that 2v needs to be multiplied by 1v² to eliminate the first term
<u> v² </u>
2v - 2 | 2v³ - 16v² + 0v + 13
<u>- (2v³ - 2v²)</u>
0 - 14v²
multiply your estimate by your divisor and subtract from the dividend.
bring down the next term and repeat.
<u> v² -7v </u>
2v - 2 | 2v³ - 16v² + 0v + 13
<u>- (2v³ - 2v²)</u>
0 - 14v² + 0v
<u>-(-14v² + 14v)</u>
- 14v
repeat again
<u />
<u> v² - 7v - 7 </u>
2v - 2 | 2v³ - 16v² + 0v + 13
<u>- (2v³ - 2v²)</u>
0 - 14v² + 0v
<u>-(-14v² + 14v)</u>
- 14v + 13
<u>-(-14v + 14)</u>
-1
and remainder gets put over the divisor and appended
v² - 7v - 7 - 1/(2v - 2)
13) Same process
<u> </u>
5a + 1 | 40a³ - 12a² - 39a - 5
<u> 8a² </u>
5a + 1 | 40a³ - 12a² - 39a - 5
<u>-(40a³ + 8a²)</u>
-20a² - 39a
<u> 8a² - 4a </u>
5a + 1 | 40a³ - 12a² - 39a - 5
<u>-(40a³ + 8a²)</u>
-20a² - 39a
-(-<u>20a² - 4a)</u>
-35a - 5
<u> 8a² - 4a - 7 </u>
5a + 1 | 40a³ - 12a² - 39a - 5
<u>-(40a³ + 8a²)</u>
-20a² - 39a
-(<u>20a² - 4a)</u>
-35a - 5
-<u>(-35a - 7)</u>
2
8a² - 4a - 7 + 2/(5a + 1)
Answer:
-1.5 per month
Step-by-step explanation:
-3 sales/2 months
