I think it would be 12.207… use the Pythagorean theorem (a^2+b^2=c^2) so 10^2+7^2
100+49=149
Then take square root of 149 to undo the the squaring of c and it should be around 12.207 :)
First I would change the descriptions of the numbers into expressions.
first number is n
second number is n + 6
third number is 4n (4 x n)
Then you would insert these expressions into an equation and isolate n.
n + n + 6 + 4n = 144
n + n + 4n = 144 - 6
6n = 138
n = 138/6
n = 23
Lastly, you would plug in this value into all of the expressions.
first number is 23
second number is 23 + 6 = 29
third number is 4(23) = 92
Therefore, the numbers are 23, 29, and 92.
Answer:
8
Step-by-step explanation:
=(7−(−1))2+(4−4)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√
=(8)2+(0)2‾‾‾‾‾‾‾‾‾‾√
=64+0‾‾‾‾‾‾√
=6‾√4
=8
Answer: i think (4n+1)^2(4n-1)^2 isnt a multiple of 8 for all integers of n because:
(4n + 1)²(4n - 1)²
= [(4n + 1)(4n - 1)]²
= (16n² - 1)²
= 16².n².n² - 2.16.n² + 1
= 8n²(32n² - 4) + 1
can see 8n²(32n² - 4) is a multiple of 8 but 1 isnt a multiple of 8
=> (4n + 1)²(4n - 1)² isnt a multiple of 8 for all integers of n.
Step-by-step explanation:
<span>Which of the following equals 140 to nearest 10
A.134
B.145
C.136
D.146</span>
