If you expand the left hand side, the inequality becomes
As you can see, appears on both sides, which means that it cancels out. You can subtract from both sides to get
and this is clearly always true. If you interpreted the inequality with words, it would sound like
"For which values of n is 18 greater than 8"
Since the inequality does not depend on n anymore and the remaining part is true (18 is indeed greater than 8), every possible value of n satisfies the inequality.
112=2∗56, I believe would be the answer to your question. (<span>The Prime Factors of 112: </span><span>2^4 • 7)</span>
Answer:
in this method a variable is expressed in terms of another variable from one equation and it is substituted in remaining equation.
Step-by-step explanation:
solve:x+2y=9 and 3x-y=13.
x+ 2y=9…be the first equation
3x-y=13…be the second equation
from first equation
x=9-2y…be the third equation substituting value of x from third equation in second equation, we get
3(9-2y)-y=13
or, 27-6y-y=13
or, y=2
now,
substituting value of y in third equation, we get,
x=9-2×2
=9-4
=5
the required value of x and y are 5&2
Answer:
n=4
Step-by-step explanation:
See the steps below:)
Answer: (a) -14 + 15n - 15, (b) 670
Step-by-step explanation:
The nth term of an Arithmetic sequence is
Tn = a + ( n - 1 )d
From the question, the 12th term = 41 and the 4th term = 1, to find the a the first term and d the common difference of the sequence, requires a little understanding. Now, we have to resolve to a simultaneous equation to get the two unknown. Now let's go:
From.the first statement
T12 = a + ( n - 1 ) = 41 --------------- 1,
Second statement,
T4 = a + ( n - 1 ) = 1 ------------------ 2
Now solve the two equations together using any known methods
T12 = a + 11d = 41
T4 = a + 3d = 1
-------------- , now subtract.
8d = 40 and
d = 5.
To find the value of a, in oder to proceed to the answers to the question, substitute for d in any of the equation above.
a + 3d = 1
a + 3 × 5 = 1
a + 15 = 1
a = 1 - 15
a = -14
So a = -14 and d = 5, Now answers to the questiona are
(a) Tn = a + ( n - 1 )d
= -14 + ( n - 1 )5
= -14 + 15n - 15
(b) sum of the 20 terms.
S20 = n/2{(2a + ( n - 1 )d )}
= 20/2{( 2× (-14) + (20 -1) × 5
= 10(-28 + 19 × 5 )
= 10( -28 + 95)
= 10( 67)
= 670.